Complex SQL 2 | find new and repeat customers | SQL Interview Questions

This video is about finding new and repeat customers .using SQL. In this video we will learn following concepts:
how to approach complex query step by step
how to use CASE WHEN with SUM
how to use common table expression (CTE)
Here is the script :
create table customer_orders (
order_id integer,
customer_id integer,
order_date date,
order_amount integer
);
select * from customer_orders
insert into customer_orders values(1,100,cast('2022-01-01' as date),2000),(2,200,cast('2022-01-01' as date),2500),(3,300,cast('2022-01-01' as date),2100)
,(4,100,cast('2022-01-02' as date),2000),(5,400,cast('2022-01-02' as date),2200),(6,500,cast('2022-01-02' as date),2700)
,(7,100,cast('2022-01-03' as date),3000),(8,400,cast('2022-01-03' as date),1000),(9,600,cast('2022-01-03' as date),3000)
;

[@satyamgoyal942]

Select a.order_date, Sum(Case when a.order_date = a.first_order_date then 1 else 0 end) as new_customer, Sum(Case when a.order_date != a.first_order_date then 1 else 0 end) as repeat_customer from( Select customer_id, order_date, min(order_date) over(partition by customer_id) as first_order_date from customer_orders) a group by a.order_date;

[@ankitbansal6]

Good.

[@tanushreesaha2276]

impressed

[@psyche5184]

A very good solution, I feel mine is overcomplicating things

[@aniket9595]

Nice one Satyam 😊

[@TonnyPodiyan]

Nice one bro

[@mananyadav6401]

Hey Ankit , your channel is really addictive. Since yesterday I have picked more than 15 problems in a row (which indirectly means I watched 15 videos from your channel straight in a row). I am really enjoying it. People binge watch Netflix and here I am binge watching sql problem solving. Can't express in words, felt like I found the gem on the KZbin. It pumped adrenaline rush in my body when I am able to solve the problems without looking at the solution. At the end comparing my solution with your provided solution and that also is teaching me how to solve any problem in minimal joins and subqueries. Amazing....Amazing...Amazing....Thank you so much for all your hardwork and knowledge sharing.

[@ankitbansal6]

Thats a big compliment for me. Keep rocking 😊

[@shankrukulkarni3234]

For me also same thing,I am not forcefully learning, by intrest I am coming and watching your videos....I can say I addicted to your channel. The positive of your channel is you will provide table with data so it make us to practice while watching your video

[@mansipuhan4827]

Same for me also

[@Tusharchitrakar]

Exactly the same here too. I'm not able to sleep at night till i solve problems and i never see your solution till i solve it myself. Thanks for this good collection of questions

[@sowmya6471]

Exactly..Same here addicted to the explaination and over enthusiastic to solve. Initially to type single line query I was seeing and typing ,now I am watching whole video till end and then writing 5-6lines queries without seeing. @AnkitBansalYou got great teaching skills.

[@BhanuGupta-x2j]

Very good question. WITH cte AS ( SELECT order_id, customer_id, order_date, order_amount, RANK() OVER(PARTITION BY customer_id ORDER BY order_date) AS rnk FROM customer_orders ) SELECT order_date, SUM(CASE WHEN rnk =1 THEN 1 ELSE 0 END) AS new_customer_count, SUM(CASE WHEN rnk 1 THEN 1 ELSE 0 END) AS repeat_customer_count FROM cte GROUP BY order_date;

[@ankitbiswas8380]

I was asked exactly the same question in my interview with dunnhumby and I failed to answer as I panicked and tried to give an answer hurriedly . Now after going through your video in steps , I completely understood the approach in how to deal with these questions. Looking forward to the rest of the playlist .

[@ankitbansal6]

Cool

[@siddhantsoni1575]

@@ankitbansal6 I was also asked this questions

[@sachinkapoor2424]

For which job position ankit

[@ankitbiswas8380]

@@sachinkapoor2424 DS

[@kartikjuneja7807]

For how many years experience did you apply?

[@PraveenSinghRathore-df3td]

Hi Ankit, 2 years I cracked the DA role with your help, now when I'm preparing again for a switch, this is my go to source material for SQL Prep, thanks for a splendid playlist. Here is my solution: SELECT order_date, sum(CASE WHEN rn2 = 1 THEN 1 ELSE 0 END) AS new, sum(CASE WHEN rn2 > 1 THEN 1 ELSE 0 END) AS repeat FROM (SELECT *, row_number() over(PARTITION BY customer_id ORDER BY order_date ASC, customer_id ASC) AS rn2 FROM customer_orders) GROUP BY order_date ORDER BY order_date;

[@swethathiruppathy9973]

Hi Sir, Thank you for all your videos ..Really helpful for learning . Here is my query with cte as (select customer_id,min(order_date) as first_visit_date from customer_orders group by customer_id) select c.order_date, sum(case when c.order_date = f.first_visit_date then 1 else 0 end) as first_visit_flag, sum(case when c.order_date != f.first_visit_date then 1 else 0 end) as repeat_visit_flag, sum(case when c.order_date = f.first_visit_date then order_amount else 0 end) as newCustAmount, sum(case when c.order_date != f.first_visit_date then order_amount else 0 end) as repeatCustAmount from customer_orders c inner join cte f on c.customer_id=f.customer_id group by c.order_date ;

[@devendrabhumarapu7869]

Assignment query: with cte as( select order_date,order_amount, row_number() over(partition by customer_id order by order_date asc) as rn from customer_orders) select order_date, sum(case when rn=1 then 1 else 0 end) as new_customers, sum(case when rn>1 then 1 else 0 end) as repeat_customers, sum(case when rn=1 then order_amount else 0 end) as new_customers_order_amount, sum(case when rn>1 then order_amount else 0 end) as repeat_customers_order_amount from cte group by order_date; select * from customer_orders;

[@yashwani209]

Now i will never Forget CTE... Great teaching skill

[@kothapalliramana4955]

MYSQL Query for the same:- with cte as( select order_date, row_number() over(partition by customer_id order by order_date asc) as rn from customer_orders) select order_date, sum(case when rn=1 then 1 else 0 end) as new_customers, sum(case when rn>1 then 1 else 0 end) as repeat_customers from cte group by order_date;

[@ankitbansal6]

This is good. Thanks for posting 👏

[@rakeshchaudhary3055]

I ended up being very close to your solution with a little difference. with old_new_counter as ( SELECT *,row_number() over (partition by customer_id order by order_date) old_new_flag FROM customer_orders) select order_date, count(case when old_new_flag=1 then 'new_customer' end) count_new_customer, count(case when old_new_flag>1 then 'old_customer' end) count_repeat_customer from old_new_counter group by order_date order by order_date; Cheers

[@ls47295]

This query will not give expected output in a case where same user has more than 1 order the same date. I tested using same records 2 times in a table . Just my input..

[@Ian15278]

Thanks, Ankit for this brainstorming question, MY QUERY SELECT order_date, Count(CASE WHEN rnk = 1 THEN cnt END) AS "New Customer", Count(CASE WHEN rnk > 1 THEN cnt END) AS "Old Customer" FROM ( SELECT order_date, customer_id, DENSE_RANK() OVER (PARTITION BY customer_id ORDER BY order_date) AS rnk, COUNT(*) OVER (PARTITION BY customer_id, order_date) AS cnt FROM customer_orders1 ) A GROUP BY order_date;

[@Nikhilg-rs7iv]

seriously no one in the entire youtube explained CTE like this. Made it so simple thank you ankit bro

[@ankitbansal6]

🙏

[@maxsteel4590]

I was asked the same question in curefit in 3rd round. There were 2 extra tables to refer but now I realize it could have been done using single table with min order date criteria. Glad I stumbled on your channel

[@ankitbansal6]

Glad to know 🙏

[@m04d10y1996]

You got this for which profile.

[@maxsteel4590]

@@m04d10y1996 Product analytics

[@gagansingh3481]

sELECT order_date, Sum(cASE WHEN ORDER_DATE = fIRST_dATE THEN 1 else 0 END) AS nEW, Sum(cASE WHEN ORDER_DATE fIRST_dATE THEN 1 else 0 END )AS rEPEATCUS FROM (Select Customer_id,order_date, Min(order_date) over (Partition by customer_id order by order_date) as First_Date from customer_orders ) as a group by order_date got the answer by this too thanks

[@arijitsaha5499]

Hi Ankit, thanks for creating such videos. Here is my approach: with sequenced_order_table as( select *, dense_rank() over(partition by customer_id order by order_date) as order_seq from customer_orders) SELECT order_date, count(case when order_seq = 1 then customer_id end) as new_customer, count(case when order_seq > 1 then customer_id end) as old_customer FROM sequenced_order_table group by 1 order by 1

[@kartikjuneja7807]

Bro , did you check this in the db?

[@Prasad1487]

MY only SQL guru.. Thank you guruji.. love you for ever.

[@susmitakundu6120]

Thanks Ankit for your guidance. Please have a look below query select sum(case when tc> 1 then 1 else 0 end )as repeat_customer, sum(case when tc= 1 then 1 else 0 end )as new_customer from (select customer_id, count( customer_id) as tc from customer_orders group by customer_id) a ;

[@shubhamchakravorty]

Thanks, ankit for this brainstorming question, initially couldnt figure out the but the min(order_date) that you gave was the key. I accomplished this with subqueries: select order_date,count(new_customer) as new,count(repeat_customer) as repeat from ( select order_date, case when order_date=first_order_date then 'new_customer' end as new_customer, case when order_date!=first_order_date then 'repeat_customer' end as repeat_customer from ( select a.*,b.first_order_date from customer_orders a join( select customer_id,min(order_date) as first_order_date from customer_orders group by customer_id) b on a.customer_id=b.customer_id)c)d group by order_date order by order_date asc;

[@tamojeetchatterjee9385]

Hey Ankit Thanks for providing this question my solution for this problem with cte as (select order_id , customer_id , order_date , lag(customer_id)over(partition by customer_id order by order_date) as statements from customer_orders) select order_date , sum(case when statements is null then 1 else 0 end) as new_customer_count , sum(case when statements is not null then 1 else 0 end) as old_customer_count from cte group by order_date order by order_date

[@vigneshshetty2149]

Hey Ankit, Thank you for this problem questions. I specially like the assignment you give at the end. As I am a beginner for SQL its very my encouraging and confidence boosting for me. Assignment sol:- with first_visit_flag as (SELECT customer_id, MIN(order_date) as first_visit_date FROM customer_orders GROUP BY customer_id), repeat_visit_flag AS ( SELECT co.order_date, fv.first_visit_date, CASE WHEN co.order_date=fv.first_visit_date THEN 1 ELSE 0 END AS first_visit, CASE WHEN co.order_date!=fv.first_visit_date THEN 1 ELSE 0 END AS repeat_visit, CASE WHEN co.order_date=fv.first_visit_date THEN SUM(order_amount) ELSE 0 END AS new_order, CASE WHEN co.order_date!=fv.first_visit_date THEN SUM(order_amount) ELSE 0 END AS repeat_order FROM customer_orders co inner join first_visit_flag fv ON co.customer_id = fv.customer_id GROUP BY co.order_date,fv.first_visit_date ) SELECT order_date, SUM(first_visit) as new_customer,SUM(repeat_visit) as repeat_customer, SUM(new_order) AS new_order_amount, SUM(repeat_order) AS repeat_order_amount FROM repeat_visit_flag GROUP BY order_date;

[@rakeshpanigrahi577]

Thanks Ankit, Here is my solution - with cte as ( select *, count(order_id) over(partition by customer_id order by order_date rows between unbounded preceding and current row) as cnt from customer_orders ) select order_date, sum(case when cnt = 1 then 1 else 0 end) as new_cust_ind, sum(case when cnt > 1 then 1 else 0 end) as repeat_cust_ind, sum(case when cnt = 1 then order_amount else 0 end) as new_cust_amt, sum(case when cnt > 1 then order_amount else 0 end) as repeat_cust_amt from cte group by order_date;

[@ft_peakhd2921]

I solved this using Window functions and cte, My solution: with cte as (select customer_id,row_number() over(partition by customer_id order by customer_id) num,order_date from customer_orders group by customer_id,order_date) select order_date, sum(case when num=1 then 1 else 0 end ) as new_cust,sum(case when num>1 then 1 else 0 end) repeat_cust from cte group by order_date;

[@rakshithareddy498]

Here is my query With cte_1 as ( Select *, rank() over(partition by customer_id order by order_date) as ranked from customer_orders ), cte_2 as ( Select order_date, case when ranked = 1 then 'new' else 'repeat' end as new_or_repeat from cte_1 ) Select order_date, sum(case when new_or_repeat = 'new' then 1 else 0 end) as new_customer,sum(case when new_or_repeat = 'repeat' then 1 else 0 end) as repeat_customer from cte_2 group by order_date; Thank you for your efforts

[@Mayank-jw9yy]

Thanks Ankit for great explanatory video> Here is solution of assignment given in video WITH first_visit AS ( SELECT customer_id, min(order_date) AS first_visit_date FROM customer_orders GROUP BY customer_id) SELECT co.order_date, SUM(CASE WHEN co.order_date = fv.first_visit_date THEN 1 ELSE 0 END) AS first_visit_customer, SUM(CASE WHEN co.order_date != fv.first_visit_date THEN 1 ELSE 0 END) AS repeat_visit_customer, SUM(CASE WHEN co.order_date = fv.first_visit_date THEN order_amount ELSE 0 END) AS first_visit_customer_order, SUM(CASE WHEN co.order_date != fv.first_visit_date THEN order_amount ELSE 0 END) AS repeat_visit_customer_order FROM customer_orders co INNER JOIN first_visit fv ON co.customer_id = fv.customer_id GROUP BY co.order_date

[@zeeshanahmed2594]

Thank you very much Sir, for this practical question and your step by step explanation.

[@ankitbansal6]

You are most welcome

[@debashreesarkar1403]

with cte1 as (select order_date, case when customer_id= rep_cs then 1 else 0 end as rep_flag, case when customer_id rep_cs then 1 else 0 end as new_flag from (select order_date,customer_id,lag(customer_id,3,0) over(order by order_id )as rep_cs from customer_orders) e1) select order_date,sum(new_flag) as new_customer, sum(rep_flag) as rep_customer from cte1 group by order_date

[@debashreesarkar1403]

using lag function with cte

[@jainwho]

Hi Ankit, with cte as (select *, row_number() over (partition by customer_id order by order_date) as order_flag from customer_orders) select order_date, sum(case when order_flag=1 then 1 else 0 end) as new_customer_count, sum(case when order_flag>1 then 1 else 0 end) as repeat_customer_count from cte group by order_date

[@suman3316]

this what a real time problems...thanks and keep bring such

[@ankitbansal6]

Sure. Thanks.

[@ujjwalvarshney3188]

select order_date, sum(case when order_date =mn then 1 else 0 end) as new_customer ,sum(case when order_date mn then 1 else 0 end) as old_customer from ( select * ,min(order_date)over(partition by customer_id) as mn from customer_orders ) group by 1 order by 1

[@bukunmiadebanjo9684]

Great content as always. Here is my attempt to the homework with first_order_table as (select customer_id, min(order_date) as first_order_date from customer_orders group by customer_id) select co.order_date, sum(case when co.order_date = fot.first_order_date then co.order_amount else 0 end) as order_amount_by_new_customer, sum(case when co.order_date fot.first_order_date then co.order_amount else 0 end) as order_amount_by_repeat_customer from customer_orders co join first_order_table fot on fot.customer_id = co.customer_id group by co.order_date order by 1

[@husnabanu4370]

Hi Ankit Thanks your videos are helping me to break down complex scnearios into smaller parts and then combine the whole query.....so i did the query in different way..Please do let me know if thats correct since the motive is only to find duplicate and new customers with cte as ( select customer_id,count(1) as ranking from customer_orders group by customer_id) select * , case when ranking>1 then 'Duplicate' else 'New' end as status_customer from cte; customer_id ranking status_customer 100 3 Duplicate 200 1 New 300 1 New 400 2 Duplicate 500 1 New 600 1 New

[@florincopaci6821]

Thank you for this video! Please come with problems like this. Thank you

[@ankitbansal6]

Sure.

[@mohdtoufique7446]

Hi Ankit..Thanks for your efforts.. I have an alternate solution as well WITH CTE AS( select *,CASE WHEN(DENSE_RANK()OVER(PARTITION BY customer_id ORDER BY order_date)=1) THEN 'New' ELSE 'Repeat' END AS IND_CUSTOMER from customer_orders ) SELECT order_date,count(CASE WHEN IND_CUSTOMER='New' THEN order_id END) AS no_new_customer, count(CASE WHEN IND_CUSTOMER='Repeat' THEN order_id END) AS no_repeat_customer FROM CTE GROUP BY order_date ORDER BY order_date

[@ankitbansal6]

Thanks for posting. This is also good. 👍

[@SundarKumar-pg6wx]

I have implemented with this logic. with cte as ( select *,ROW_NUMBER() over(partition by customer_id order by order_date asc) as rnk from customer_orders), cte_not_1 as (select order_date,count(*) as cnt from cte where rnk 1 group by order_date) select * from ( select order_date,count(*) as new_customer_count from cte t1 where t1.rnk = 1 group by order_date) t1 left join cte_not_1 t2 on t1.order_date = t2.order_date

[@anishchhabra6085]

Solved the question without looking into the solution in MySQL, I have used the concept of sum with case when after seeing it in your other video, it is very helpful and important concept: with cte as ( select *, row_number() over(partition by customer_id) as rn from customer_orders order by order_date ) select order_date, sum(case when rn = 1 then 1 else 0 end) as new_customer_count, sum(case when rn = 1 then 0 else 1 end) as repeat_customer_count from cte group by order_date; Please let me know if there is some issue in this code

[@techlearnersmarathi5461]

Thanks for sharing this problem and approach behind same. below is my query select co.order_date, sum(case when CO.order_date= fv.first_date then 1 else 0 end) as no_of_new_cust, sum(case when CO.order_date!= fv.first_date then 1 else 0 end) as no_of_repeat_cust, sum(case when CO.order_date= fv.first_date then order_amount else 0 end) as total_amount_by_new_cust, sum(case when CO.order_date!= fv.first_date then order_amount else 0 end) as total_amount_by_old_cust from [dbo].[customer_orders] co inner join (select customer_id, min(order_date) as first_date from [dbo].[customer_orders] group by customer_id) fv on co.customer_id=fv.customer_id group by order_date order by CO.order_date;

[@parth_pm16]

Hi everyone, HW Task: add two columns of first_visit_order_amount, last_first_visit_order_amount Solution: sum(case when fv.first_visit_date = co.order_date then co.order_amount else 0 end) as first_visit_Order_amt_flag , sum(case when fv.first_visit_date != co.order_date then co.order_amount else 0 end) as repeat_visit_order_amt_flag add this two columns in Ankit's solution.

[@abhishekjain4895]

Hey Ankit, I have used a different approach: with new_table as( select order_date,count(customer_id) as new_customer from customer_orders a where 0=( select count(*) from customer_orders b where a.order_date>b.order_date and a.customer_id=b.customer_id) group by order_date), repeat_table as (select order_date,count(customer_id) as old_customer from customer_orders a where ( select count(*) from customer_orders b where a.order_date>b.order_date and a.customer_id=b.customer_id)>0 group by order_date) select case when a.order_date is null then b.order_date else a.order_date end as date ,new_customer,old_customer from new_table a full outer join repeat_table b on a.order_date=b.order_date;

[@saib7231]

hi ankith this is also working with cte as ( SELECT *,min(order_date) over (partition by customer_id) as first_date FROM customer_orders as a ) select order_date,count(case when order_date first_date then customer_id end) as repeat, count(case when order_date = first_date then customer_id end) as new, count(customer_id) as total from cte group by order_date

[@user-yk4xd4gk6o]

Hi Ankit,your channel is very helpful and the way you are explaining is just amazing. Here is my solution for this,WITH CTE AS ( SELECT ORDER_ID,CUSTOMER_ID,ORDER_DATE,ORDER_AMOUNT, CASE WHEN PRIV IS NULL THEN 1 ELSE 0 END AS NEW_FLAG, CASE WHEN PRIV IS NOT NULL THEN 1 ELSE 0 END AS OLD_FLAG FROM ( select *, lag(ORDER_DATE) over(partition by CUSTOMER_ID order by ORDER_DATE) as PRIV from customer_orders) ORDER BY ORDER_ID) SELECT ORDER_DATE,SUM(NEW_FLAG) AS NEW_CUSTOMER,SUM(OLD_FLAG) AS OLD_CUTOMER FROM CTE GROUP BY ORDER_DATE;

[@MyTetere]

This is a clear and concise explanation

[@sarveshrajan88]

Amazing Video Ankit, Commenting my approach SELECT B.order_date, SUM(CASE WHEN FC.customer_id IS NULL THEN 0 ELSE 1 END) AS NEW_CUSTOMER_COUNT ,SUM(CASE WHEN FC.customer_id IS NULL THEN 1 ELSE 0 END) AS REPEAT_CUSTOMER_COUN ,SUM(CASE WHEN FC.CUSTOMER_ID IS NOT NULL THEN B.ORDER_AMOUNT ELSE 0 END) AS NEW_CUSTOMER_SUM ,SUM(CASE WHEN FC.customer_id IS NULL THEN B.order_amount ELSE 0 END) AS REPEAT_CUSTOMER_SUM FROM customer_orders AS B LEFT JOIN (SELECT A.CUSTOMER_ID,A.order_date,ROW_NUMBER() OVER(PARTITION BY A.CUSTOMER_ID ORDER BY A.ORDER_DATE) AS RNK FROM customer_orders AS A) AS FC ON B.order_date=FC.order_date AND B.customer_id=FC.customer_id AND FC.RNK='1' GROUP BY B.order_date

[@plumbarch]

There is a small difference between the repeat customers (those purchased for the consecutive days) and old customers (no consecutive days condition). For the repeat customers - purchasing for consecutive days select order_date, sum(new_customer) as new_customer, sum(repeat_customer) as repeat_customer from (select customer_id, order_date, case when order_date=first_visit then 1 else 0 end as new_customer, case when date_diff(order_date, prev_day)=1 then 1 else 0 end as repeat_customer ( select customer_id, order_date, min(order_date) over (partition by customer_id) as first_visit , lag(order_date,1,0) over (partition by customer_id order by order_date) as prev_day from customer_orders)x)y group by order_date

[@bapanbairagya9642]

awesome problem, Thank you so much for posting.

[@techmania979]

I used below query, with cte as( select *,min(order_date) over(partition by customer_id) as first_visit from customer_orders ) select order_date, sum( case when order_date=first_visit then 1 else 0 end ) as first_time, sum( case when order_datefirst_visit then 1 else 0 end ) as sec_time from cte group by order_date

[@ankitbansal6]

Thank you for posting.

[@Ashu23200]

My assignment query with cte as ( select customer_id, order_date, order_amount, MIN(order_date) over (partition by customer_id) as first_order_date from customer_orders ) , cte2 as ( select *, case when order_date = first_order_date then 'New_Customer' else 'Repeat_Customer' end as Customer_Type --, COUNT(1) as mnjh from cte ) select Customer_Type,COUNT(1) as Count_of_Customers ,sum(order_amount) as sales from cte2 group by Customer_Type

[@arthurmorgan9010]

My solution was: with firsttable(customerid,firstorderdate) as ( select customer_id,min(order_date) as first_order_date from customer_orders group by customer_id ) select order_date,sum(case when (customerid = customer_id and firstorderdate = order_date) then 1 else 0 end) as newcustomer, sum(case when (customerid = customer_id and firstorderdate != order_date) then 1 else 0 end) as oldcustomer from customer_orders,firsttable group by order_date

[@smitpatel8782]

select c.order_date , sum(case when c.customer_id in (select co.customer_id from customer_orders co where co.order_date < c.order_date) then 1 else 0 end) as Repeat_customer, sum(case when c.customer_id not in (select co.customer_id from customer_orders co where co.order_date < c.order_date) then 1 else 0 end) as Not_Repeat_customer from customer_orders c group by 1

[@bhajans-m3l]

with cte as ( select *, row_number () over (partition by customer_id order by order_date ) as rn from customer_orders), cte1 as (select order_date , count(1) as new_cust from cte where rn=1 group by order_date), cte2 as ( select order_date , count(1) as rep_cust from cte where rn>1 group by order_date) select a.order_date,new_cust, case when rep_cust is null then 0 else rep_cust end as rep_cust from cte1 a left join cte2 b on a.order_date=b.order_date

[@GanesanDinesh]

Hi Ankit, Thanks for your effort ;with cte_1 as (Select customer_id,order_date, row_number() over(partition by customer_id order by min(order_date)) as first_order_date from customer_orders group by customer_id,order_date ) select order_date, sum(case when first_order_date = 1 then 1 else 0 end) as first_visit, sum(case when first_order_date > 1 then 1 else 0 end )as repeated_visit from cte_1 group by order_date

[@ankitbansal6]

Looks good. Keep rocking 😊

[@rahuldave6699]

My solution with cte as( select *, rank() over (partition by customer_id order by order_date) as rnk from customer_orders) select order_date ,count(case when rnk = 1 then 1 else Null end ) as new_customer, count(case when rnk>1 then 1 else Null end) as repeat_customer from cte group by order_date

[@IndianHacker-hisBest]

Really good channel and informative videos.

[@ankitbansal6]

Glad you like them!

[@shubhamgoyal5227]

GREAT QUESTION ANKIT SIR HERE IS MY APPROACH: WITH CTE AS( SELECT * , ROW_NUMBER()OVER(PARTITION BY CUSTOMER_ID ORDER BY ORDER_DATE)RN FROM customer_orders) ,CTE_2 AS( SELECT *, CASE WHEN RN=1 THEN 1 ELSE NULL END AS NEW_CUST, CASE WHEN RN>1 THEN 1 ELSE NULL END AS REPEATED, CASE WHEN RN=1 THEN order_amount ELSE NULL END NEW_CUST_AMOUNT, CASE WHEN RN>1 THEN order_amount ELSE NULL END REPEATED_CUST_AMOUNT FROM CTE) SELECT order_date, COUNT(NEW_CUST) NEW_CUSTOMER_COUNT, COUNT(REPEATED) REPEATED_CUSTOMER_COUNT ,ISNULL(SUM(NEW_CUST_AMOUNT),0) AMOUNT_FROM_NEW_CUSTOMER ,ISNULL(SUM(REPEATED_CUST_AMOUNT),0)AMOUNT_FROM_REPEATED_CUSTOMER FROM CTE_2 GROUP BY order_date

[@user-ok8ou9ro2y]

with CTE as ( select *, dense_rank() over(partition by customer_id order by order_date) as rnk from customer_orders )select order_date,sum(case when rnk=1 then 1 else 0 end ) as new_order, sum(case when rnk1 then 1 else 0 end ) as repaet_customer from CTE group by order_date

[@pragatiaggarwal8103]

You are awesome 💌

[@ashish-blessings]

Thank you so much Ankit Bansal. This is really helpful.

[@shourya4092]

What a Explanation mind blowing ❤️❤️❤️

[@ankitbansal6]

Thank you so much 😀

[@ajeshrajan8079]

Great explanation for both approach and solution

[@ankitbansal6]

Glad you liked it

[@KoushikT]

Without using join *********************** with A as (select customer_id,order_date,lag(order_date) over (partition by customer_id) as previous_visit from customer_orders) select order_date, sum(case when previous_visit is null then 1 else 0 end) as new_customer, count(*)-sum(case when previous_visit is null then 1 else 0 end) as repeat_customer from A group by order_date order by order_date

[@ankitbansal6]

Good one

[@akashkamble9902]

Amazing Video and Interpretation, thanks a lot for making this video, here is the answer for assignment with fv as ( select CUSTOMER_ID, min(ORDER_DATE) as fisrtVist from customer_orders group by CUSTOMER_ID), final as( select co.ORDER_DATE, SUM(case when co.ORDER_DATE = fv.fisrtVist then 1 else 0 end) as firstVistFinal, SUM(case when co.ORDER_DATE != fv.fisrtVist then 1 else 0 end) as repeatVistFinal, SUM(case when co.ORDER_DATE = fv.fisrtVist then order_amount else 0 end) as firstVistAmout, SUM(case when co.ORDER_DATE != fv.fisrtVist then order_amount else 0 end) as repeatVistAmout from customer_orders co , fv where fv.CUSTOMER_ID = co.CUSTOMER_ID group by co.ORDER_DATE ) select * from final

[@debabratabar2008]

Hi ankit , added solution for assignment part with customer_first_ord_dt as ( select customer_id , min( order_date) as first_order_date from customer_orders group by 1 ) , final_temp_table as ( select * , case when order_date = first_order_date then 1 else 0 end as new_customer_count , case when order_date != first_order_date then 1 else 0 end as repeat_customer_count from customer_orders co inner join customer_first_ord_dt as cot on co.customer_id = cot.customer_id ) select order_date , sum( case when order_date = first_order_date then 1 else 0 end) as new_customer_count , sum( case when order_date != first_order_date then 1 else 0 end) as repeat_customer_count , sum( case when new_customer_count = 1 then order_amount else 0 end) as new_customer_sales, sum( case when repeat_customer_count = 1 then order_amount else 0 end) as repeat_customer_sales from final_temp_table group by 1 order by 1

[@janhavighuge7095]

Thankyou for another great question! My solution to this question: SELECT new.order_date, COUNT(CASE WHEN previous.customer_id IS NULL THEN 1 END) AS new_cust, COUNT(DISTINCT(CASE WHEN previous.customer_id IS NOT NULL THEN previous.customer_id END)) AS repeat_cust, SUM(CASE WHEN previous.customer_id IS NULL THEN new.order_amount END) AS amount_by_new_cust, SUM(DISTINCT(CASE WHEN previous.customer_id IS NOT NULL THEN new.order_amount ELSE 0 END)) AS amount_by_repeat_cust FROM customer_orders new LEFT JOIN customer_orders previous ON previous.customer_id = new.customer_id AND previous.order_date < new.order_date GROUP BY new.order_date;

[@Brown_Munda_oo7]

VERY Good Query

[@keerthanakalyanapu7475]

Select a.order_date, Sum(Case when a.order_date = a.first_order_date then 1 else 0 end) as new_customer, Sum(Case when a.order_date != a.first_order_date then 1 else 0 end) as repeat_customer, Sum(Case when a.order_date = a.first_order_date then A.order_amount else 0 end) as new_customer_amt, Sum(Case when a.order_date != a.first_order_date then A.order_amount else 0 end) as repeat_customer_amt from( Select customer_id, order_date,order_amount, min(order_date) over(partition by customer_id) as first_order_date from customer_orders) a group by a.order_date order by a.order_date;

[@vaibhavtiwari8670]

# 1 asigmnet solution case when first_visit_date=order_date then order_amount else 0 end as amount_spend_new, case when first_visit_date!=order_date then order_amount else 0 end as amount_spend_old select sum(amount_spend_new) as new_customer_spend,sum(amount_spend_old) as old_customer_spend,sum(new_customer) as new_customers,sum(repeat_customer) as repeat_customer,order_date from customer group by order_date assignment completed

[@kunnalkhatreja6191]

select b.order_date,sum(case when a.first_order = b.order_date then 1 else 0 end) as first_time_customers, Sum(case when a.first_order b.order_date then 1 else 0 end) as repeat_customers, sum(case when a.first_order = b.order_date then order_amount end) as new_cust_order,sum(case when a.first_order b.order_date then order_amount end) as repeat_cust_order from (select customer_id,min(order_date) as first_order from customer_orders group by customer_id) a join customer_orders b on a.customer_id=b.customer_id group by b.order_date

[@saurabhsomkuwar1276]

Wow Ankit, your videos on SQL are so good, informative and helpful. Thanks a lot for making them. Keep going.

[@ankitbansal6]

Thank you 😊

[@sandeepanmahapatra4888]

This was my approach, using RANK() concept - WITH Orders_with_rank AS ( SELECT *, ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY order_date) AS Order_rank, CASE WHEN Order_rank = 1 THEN 1 ELSE 0 END AS "New Customer" , CASE WHEN Order_rank > 1 THEN 1 ELSE 0 END AS "Returning Customer" FROM customer_orders ) SELECT order_date, SUM("New Customer") AS "New Customers", SUM("Returning Customer") AS "Returning Customers" FROM Orders_with_rank GROUP BY order_date

[@ashwingupta4765]

Query with Assignment with cte as ( select *, min(order_date) over(partition by customer_id) as first_vist_date from customer_orders ) Select order_date, sum(case when order_date = first_vist_date then 1 else 0 end) as first_customer, sum(case when order_date!=first_vist_date then 1 else 0 end) as repeat_customer, sum(case when order_date = first_vist_date then order_amount else 0 end) as first_customer_amount, sum(case when order_date!=first_vist_date then order_amount else 0 end) as repeat_customer_amount from cte group by order_date

[@vijaygupta7059]

using other method with cte as( Select * , case when customer_id in (select customer_id from customer_orders where order_date

[@dfkgjdflkg]

love your great content.

[@akashwatar6633]

I am planning to complete all the SQL videos created by you in order to learn SQL. I will post a comment on each video and like it as a checklist for completed videos, starting from the beginning.

[@nidhisingh4973]

----------------count of new and repeat customer with cte as( select customer_id, order_date,case when order_date =min(order_date) over (partition by customer_id) then 'New customer' end as new, case when order_date min(order_date) over (partition by customer_id) then 'Repeat' end as r from customer_orders) select distinct order_date, count(new) as new_customer ,count(r) as repeat_customer from cte group by order_date

[@md.shabbirhossainbhuiyea4363]

What if the same customer visits the website twice or thrice and orders each time? In that case, he should be a repeat customer. However, according to your solution, he won't be counted as a repeated customer as his min(order_date) = order_date. What do you think? However, your tutorials have been really helpful to me. Really appreciate your effort.

[@mdfurqan]

Hey so what do you think the solution for this ? Could you please help in this ?

[@vandanaK-mh9zo]

then I think we need timestamp as well not only the date part.

[@mocococo2877]

Well, he will be both new AND repeat customer as per current problem explanation. So in this case new problem explanation will be needed.

[@user-ze4xc6tu4f]

best step by step practice

[@prashant887]

with temp as ( select cust_id,order_date,lag(order_date) over(PARTITION by cust_id order by order_date) as prev_order from cust_orders ) select trunc(order_date), sum(case when prev_order is null then 1 else 0 end) first,sum(case when prev_order is null then 0 else 1 end) as prev from temp group by trunc(order_date);

[@ankitbansal6]

Thanks for posting 👏

[@harshithapolisetty3501]

with cte as(select order_id,customer_id,order_date,order_amount, row_number() over (partition by customer_id order by order_Date) as rk from customer_orders ) select order_date, sum(case when rk=1 then 1 else 0 end) as new, sum(case when rk>1 then 1 else 0 end) as repeat from cte group by order_date

[@gauravgupta5530]

What a beautiful question, make your brain to hit hard.

[@kinzorize]

correct query in case you are facing error in creating the table : CREATE TABLE customer_orders ( order_id integer, customer_id integer, order_date date, order_amount integer ); INSERT INTO customer_orders VALUES (1, 100, '2022-01-01', 2000), (2, 200, '2022-01-01', 2500), (3, 300, '2022-01-01', 2100), (4, 100, '2022-01-02', 2000), (5, 400, '2022-01-02', 2200), (6, 500, '2022-01-02', 2700), (7, 100, '2022-01-03', 3000), (8, 400, '2022-01-03', 1000), (9, 600, '2022-01-03', 3000);

[@vinaykumar-hb7rf]

Love the way of explanation with step by step. 😀

[@ankitbansal6]

🙏

[@vijaypalmanit]

You make it so easy, superb explanation

[@suriyas6338]

Hi Aniket, Here is my solution :) with cte1 as( select *, ROW_NUMBER() OVER(PARTITION BY customer_id order by order_date) as row_num from customer_orders ) select order_date, sum(case when row_num = 1 then 1 else 0 end) as new_cus, sum(case when row_num > 1 then 1 else 0 end) as rep_cus, sum(case when row_num =1 then order_amount else 0 end) as newCus_order_amnt, sum(case when row_num >1 then order_amount else 0 end) as repCus_order_amnt from cte1 group by order_date

[@Mahi_RSV]

New and repeat customers each day: with cte_1 as ( select customer_id,order_date,rank() over(partition by customer_id order by order_date) as rnk from customer_orders ) select order_date, sum(case when rnk=1 then 1 else 0 end ) as new_customers, sum(case when rnk>1 then 1 else 0 end ) as repeat_customers from cte_1 group by order_date Sales by New and repeat customers each day: with cte_1 as ( select customer_id,order_date,order_amount,rank() over(partition by customer_id order by order_date) as rnk from customer_orders ) select order_date, sum(case when rnk=1 then order_amount else 0 end ) as new_customers_sales, sum(case when rnk>1 then order_amount else 0 end ) as repeat_customers_sales from cte_1 group by order_date Enjoyed solving this question! Thanks Ankit Bhai :)

[@prekshasharma1216]

Assignment: with CTE as( select customer_id, min(order_date) as first_visit from customer_orders group by customer_id) select c.order_date, sum(case when c.order_date = t.first_visit then 1 else 0 end) as first_visit_flag, sum(case when c.order_date != t.first_visit then 1 else 0 end) as repeat_visit_flag, sum(case when c.order_date = t.first_visit then order_amount else 0 end) as first_visit_amount, sum(case when c.order_date != t.first_visit then order_amount else 0 end) as repeat_visit_amount from customer_orders c join CTE t on c.customer_id = t.customer_id group by order_date order by order_date

[@venkataram6460]

Nice work Ankit, your way of solving the problem is simple but effective.

[@RajneeshDabral]

with cte as ( select order_date,status,count(*) as flag from ( select *,case when rnk in(1) then 'New' else 'Repeat' end as 'status' from ( select *,ROW_NUMBER()over(partition by customer_id order by order_date) as rnk from customer_orders) f ) g group by order_date,status ),cte2 as ( select order_date,sum(case when status = 'New' then flag end) as 'New', sum(case when status = 'Repeat' then flag end) as 'Repeat' from cte group by order_date) select order_date,ISNULL(New,0) as 'New',ISNULL(Repeat,0) as 'Repeat'from cte2

[@AbhishekSabharwal131997]

with cte as ( select customer_id,order_date, row_number() over (partition by customer_id order by order_date) as rw from customer_orders ) select order_date, sum(case when rw =1 then 1 else 0 end) as Newcust, sum(case when rw>1 then 1 else 0 end) as old from cte group by 1

[@ayushtyagi4240]

My sol : with temp as (select * , row_number() over (partition by customer_id ) visits from customer_orders order by order_date ) , temp2 as ( select order_date , case when visits = 1 then 1 else 0 end as fresh , case when visits != 1 then 1 else 0 end as repeated from temp ) select order_date , sum(fresh) new_customers , sum(repeated) repeat_customers from temp2 group by 1 order by 1

[@priyanikalje-fg4mw]

Today exactly same question was asked to Me for Cummins.

[@dakshbhatnagar]

with cte as ( select *, case when row_number() over(partition by customer_id order by order_date) = 1 then 'new' else 'repeat' end as flag from customer_orders) select order_date, flag, count(flag) as Count from cte group by order_date, flag order by order_date

[@harshSingh-if4zb]

with cte as( select distinct cust_id, first_value(order_dt) over(partition by cust_id order by order_dt) as fv from emp ) select order_dt, sum(case when c.fv = e.order_dt then 1 else 0 end ) as first_visit ,sum(case when c.fv e.order_dt then 1 else 0 end ) as repete_visit from emp e join cte c on e.cust_id = c.cust_id group by e.order_dt

[@mahenpriy]

awesome explanation ...

[@ramyagaraga3791]

Assignment Query: select A.order_date,sum(case when A.rn =1 then 1 else 0 end) as new_custmer_count, sum(case when A.rn >1 then 1 else 0 end) as repeat_customer_count, sum(case when A.rn =1 then A.ORDER_AMOUNT else 0 end) as sum_new_customer_order_amount, sum(case when A.rn >1 then A.ORDER_AMOUNT else 0 end) as sum_repeat_customer_order_amount from (select customer_id,order_date,order_amount, row_number() over(partition by customer_id order by order_date asc) as rn from customer_orders)A group by A.order_date;

[@muhammedjunaisk4835]

select order_date,sum(new_cust_i) as new_csutomers,(count(customer_id)-sum(new_cust_i)) as old_customers from ( select *,new_cust_i=case when customer_id not in (select customer_id from customer_orders c2 where c2.order_date

[@user-kg1tt8pw4t]

with temp as( select order_date, case when customer_id in (select customer_id from customer_orders c2 where c2.order_date

[@varunanr5433]

my approach - select t.order_date, t.New_Customer_Count, coalesce((b.repeated_customer_count),0) as Repeated_Customer_Count from (select order_date, count(*) as New_Customer_Count from (select *, row_number() over(partition by customer_id) as rn from customer_orders) as new_customer where rn = 1 Group by order_date) as t left join (select order_date, count(*) as repeated_customer_count from (select *, row_number() over(partition by customer_id) as rn from customer_orders) as new_customer where rn > 1 Group by order_date) as b on t.order_date=b.order_date

[@DriveWhisperer]

SELECT order_date, SUM(CASE WHEN rn = 1 THEN 1 ELSE 0 END) first_customer, SUM(CASE WHEN rn > 1 THEN 1 ELSE 0 END) repeated_customer FROM ( SELECT customer_id, order_date, ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY order_date) AS rn FROM customer_orders ) t GROUP BY order_date

[@gandlapentasabjan9115]

Task output: with cte as (select customer_id,order_date,order_amount,min(order_date) over (partition by customer_id order by order_date) as first_visit_date from customer_orders) select order_date ,sum(case when order_date=first_visit_date then order_amount end) as new_customer_order_amount ,sum(case when order_date!=first_visit_date then order_amount end) as Repeast_customer_order_amount ,sum(case when order_date=first_visit_date then 1 else 0 end) as new_customer ,sum(case when order_date!=first_visit_date then 1 else 0 end) as repeat_customer from cte group by order_date;

[@sajanpradhan1539]

such a great explanation!!!

[@ayushsrivastav8220]

with cte1 as( select *, row_number() over(partition by customer_id order by order_date) as rn from customer_orders) select order_date, count(case when rn=1 then 1 else null end) as new_orders, count(case when rn!=1 then 1 else null end) as repeat_orders from cte1 group by order_date

[@sumanacharya461]

my solution - with rp as (SELECT customer_id,order_date, rank() over (partition by customer_id order by order_date) as rn FROM customer_orders order by order_date) select order_date, sum(case when rn = 1 then 1 else 0 end) as new_customer,sum(case when rn > 1 then 1 else 0 end) as old_customer from rp group by 1 order by 1