Give me 1000 likes on this video and I will create a video on how to create a calendar table from scratch 😊
@shipra9597 ай бұрын
Bring only records with StudentMarks greater than 75. Catch is DO NOT use WHERE/GROUPBY CLAUSE Sample Input: StudentId StudentMarks 1 99 2 76 3 71 4 50 5 76 Expected Output: StudentId StudentMarks 1 99 2 76 5 76 recently i faced this qus in ey interview for data engineer with 4 yr of exp
@DEwithDhairy7 ай бұрын
PySpark Version of this problem : kzbin.info/www/bejne/mWqXmI1ug8mAfqM
@vinil92122 ай бұрын
with cte as ( select sku_id,price_date,price, ROW_NUMBER() OVER(PARTITION BY month(price_date) order by price_date desc) AS rnk FROM sku), cte2 as( SELECT sku_id,DATETRUNC(month,DATEADD(month,1,price_date)) AS next_month,price FROM cte WHERE rnk=1 UNION ALL SELECT * FROM sku WHERE DATEPART(day,price_date)=1 ) SELECT *,coalesce(price-LAG(price) OVER ( ORDER BY next_month),0) AS price_diff FROM cte2 ORDER BY next_month;
@naveenbhandari50977 ай бұрын
Hi Ankit bhai, Today I have completed all the videos from your channel. Here I just want to thank you for making such amazing videos. Your way of explaining things is really commendable, I have failed in many interviews bcos of advanced SQL concepts but this time I have gained confidence I never had. Once again thank you for making such life-changing videos. Keep doing great! may god give you all the success you wish! Thanks, Man. looking forward a great learning ahead from your channel.
@ankitbansal67 ай бұрын
Glad to know that ☺️ keep rocking 💪
@avi80167 ай бұрын
Wow, this was great 💯 I guess I'll need to work on the date function Thankyou 🙏
@girishpv81937 ай бұрын
Very good explanation Ankit... Initially I thought this looks simple..but the way you generalized the query is awesome.. Keep going 👏
@ankitbansal67 ай бұрын
Thanks a ton🙏
@addhyasumitra907 күн бұрын
my approach: following 1st method with lead-lag with CTE as ( select *, ROW_NUMBER() OVER(partition by sku_id, month(price_date) order by price_date desc) as rn from SKU) select sku_id, price_date, price from SKU where DATEPART(DAY, price_date)=1 UNION ALL select sku_id, datetrunc(month,isnull(LEAD(price_date) OVER(partition by sku_id order by price_date), DATEADD(month,1,price_date))) as next_month, price from CTE where rn=1
@rajkumarrajan80597 ай бұрын
Ankit, Please create the calendar table from scratch !!!!!
@grim_rreaperr7 ай бұрын
DECLARE @StartDate DATE = CAST('2000-01-01' AS DATE); /*set start date*/ DECLARE @EndDate DATE = CAST('2024-12-31' AS DATE); /* set end date */ WITH calendar AS ( SELECT @StartDate AS cal_dates UNION ALL SELECT DATEADD(DAY, 1, cal_dates) FROM calendar WHERE cal_dates < @EndDate ) SELECT cal_dates, MONTH(cal_dates) AS cal_month, DATEPART(DAYOFYEAR, cal_dates) AS cal_year_day, DAY(cal_dates) AS cal_month_day, DATEPART(WEEK, cal_dates) AS cal_week, DATEPART(WEEKDAY, cal_dates) AS cal_week_day, DATEPART(QUARTER, cal_dates) AS cal_quarter_num FROM calendar OPTION(MAXRECURSION 0);
@TY-zl1vw7 ай бұрын
Very informative, first time I heard about the DATE_TRUNC function, but it's not available for me to practice, since I'm using SQL Server 2018. Edit: Could DATEADD(DAY, 1,EOMONTH(price_date,0)) achieve the same ?
@vishnugottipati93736 ай бұрын
thank you
@gautamigaikwad45497 ай бұрын
Please create a video on how to create calendar table 15:10
@grim_rreaperr7 ай бұрын
DECLARE @StartDate DATE = CAST('2000-01-01' AS DATE); /*set start date*/ DECLARE @EndDate DATE = CAST('2024-12-31' AS DATE); /* set end date */ WITH calendar AS ( SELECT @StartDate AS cal_dates UNION ALL SELECT DATEADD(DAY, 1, cal_dates) FROM calendar WHERE cal_dates < @EndDate ) SELECT cal_dates, MONTH(cal_dates) AS cal_month, DATEPART(DAYOFYEAR, cal_dates) AS cal_year_day, DAY(cal_dates) AS cal_month_day, DATEPART(WEEK, cal_dates) AS cal_week, DATEPART(WEEKDAY, cal_dates) AS cal_week_day, DATEPART(QUARTER, cal_dates) AS cal_quarter_num FROM calendar OPTION(MAXRECURSION 0);
@grim_rreaperr7 ай бұрын
working as intended in ms sql server, we can other attribute columns as well like month name, day name etc
@gautamigaikwad45497 ай бұрын
Thank you
@grim_rreaperr7 ай бұрын
@@gautamigaikwad4549 🙏
@its_anky5382Ай бұрын
with cte as ( Select *,row_number() over(partition by sku_id, year(price_date),month(price_date) order by sku_id,month(price_date) asc,price_date desc) rnk from sku ) select sku_id,case when day(price_date)=1 then price_date else DATETRUNC(month,price_date) end as start_date , isnull(lag(price) over(order by price_date),price) price_at_month_start from cte where rnk=1
@nidhisingh49735 ай бұрын
Hello Ankit, Really grateful to you for all these amazing videos.
@sonalijain9024Ай бұрын
I USED THIS APPROACH SO IT WILL WORK FINE ALL POSIBILITIES. IT WILL BE A GENERIC SOLUTION .PLZ TELL ME with cte as ( select *, ROW_NUMBER() over(partition by sku_id,year(price_date),month(price_date) order by price_date desc) rnk, datetrunc(MONTH,DATEADD(month,1,price_date)) next_month from sku ) select sku_id,next_month,price from cte where rnk =1 union all select sku_id,price_date,price from sku where DATEPART(day,price_date) = 1 order by next_month asc
@akashgoel601Ай бұрын
thanks for this, posting my sol. little different approach(actually it similar to your second soln, after watching complete video i realised): with cte as ( select price_date,price, lead(DATEADD(day,-1,price_date),1,DATEADD(month,1,price_date)) OVER(order by price_date) as lead from sku ), cte_2 as ( SELECT price_date ,DATEADD(DAY, 1, EOMONTH(price_date, 0)) as frst_Date from sku ) select price_date,price from sku where DATEPART(day,price_date)=1 union all select distinct s.frst_Date,a.price from cte a join cte_2 s on s.frst_Date BETWEEN a.price_date and a.lead
@Dhanushts-g7x7 ай бұрын
with recursive cte1 as (Select min(price_date) pd from sku union all select date_add(pd,interval 1 day) pd from cte1 where pd
@KisaanTuber7 ай бұрын
Hi Ankit. Thanks for posting & explaining such challenging SQL problems. Here is my stab at the problem without using calendar table: with RECURSIVE t1 as ( SELECT date_trunc('month', MIN(price_date)) as month_date from sku UNION ALL SELECT month_date+interval '1 month' as month_date from t1 where month_date=sku.price_date) SELECT month_date, month_price from t2 where price_rnk=1 ORDER by 1;
@srikarrar2617 ай бұрын
14:24 Hi Sir, may I know what will happen instead of taking UNION ALL with UNION. I think we don't need to use Subquery to filter out the price data having 1st day of month
@kadagaladurgesh36917 ай бұрын
Great explanation Thanks for the video, I have a doubt At time 13:56 to avoid duplicates we use new condition with and operator, can we achieve same result with Union instead of union all
@sandhyakumari82047 ай бұрын
Thanks Ankit, it will be helpful if you can create a video on making of calendar table!
@ankitbansal67 ай бұрын
Okay sure
@arpanscreations69542 ай бұрын
Hi Ankit, I think we can add the condition in the cte "where day(price_date) 1" isn't it?
@apna96567 ай бұрын
Hi Ankit, It would be helpful for us, if you can create a video on calender table
@ankitbansal67 ай бұрын
Okay sure
@xiamojq6217 ай бұрын
Good video but have you thing about procedures and functions questions there are very rare in KZbin
@Amulya8697 ай бұрын
This KZbin channel is more useful.Give me some more like this
@rohitsharma-mg7hd3 ай бұрын
my simple solution: with cte1 as (SELECT month::date FROM generate_series('2023-01-01', '2024-01-01', INTERVAL '1 Month') month), cte2 as (select *,lead(price_date,1,'2023-05-01') over() as prev_date from sku) select * from cte1 c1 join cte2 c2 on c1.month between c2.price_date and c2.prev_date
@karangupta_DE7 ай бұрын
with recursive cte as ( select (select min(price_date) from sku) as all_dates union all select all_dates + interval '1 day' from cte where true and all_dates
@apurvasaraf58287 ай бұрын
with cte as (select *,RANK() over(partition by sku_id,month(price_date) order by day(price_date) desc) as r from sku) select price_date,price from sku where day(price_date)=1 union all select datetrunc(MONTH,DATEADD(month,1,price_date)) as d, price from cte where r=1
@ritusantra86417 ай бұрын
with cte as (select *, cast (dateadd(mm,DATEDIFF(mm,0,price_date)+1,0) as date) as date ,rank() over(partition by year(price_date),month(price_date) order by day(price_date) desc) as rnk from sku) select sku_id, date, price from cte where rnk = 1 union all select sku_id, price_date, price from cte where day(price_date) = 1 and month(price_date) =1 order by date;
@hariikrishnan7 ай бұрын
How many YOE candidates can expect such questions ? Found it quite hard as a fresher (< 1YoE)
@ankitbansal67 ай бұрын
It's a tough one ..4 plus YOE
@PraveenSinghRathore-df3td22 сағат бұрын
MYSQL solution: with recursive cte as ( select min(price_date) as first_date, '2023-12-31' as end_date from sku union all select date_add(first_date,interval 1 day) as first_date, end_date from cte where first_date < end_date ), cte2 as (select first_date, day(first_date) as cal_day, month(first_date) as cal_month from cte) select s.sku_id,c.first_date,s.price from cte2 c join (select sku_id,price_date,date_add(lead(price_date,1,date_add(price_date,interval 1 month)) over(partition by sku_id order by price_date),interval -1 day) as valid_till,price from sku) s on c.first_date between s.price_date and s.valid_till where cal_day = 1 order by first_date;
@sauravverma83557 ай бұрын
@ankit bansal select *,ifnull(price-lag(price,1) over(partition by sku_id),0) as Diff from (WITH cte AS ( SELECT *, ROW_NUMBER() OVER (PARTITION BY sku_id, YEAR(price_date), MONTH(price_date) ORDER BY price_date DESC) AS rn FROM sku) SELECT sku_id, STR_TO_DATE(DATE_FORMAT(price_date + INTERVAL 1 MONTH, '%Y-%m-01'), '%Y-%m-%d') AS 'date', price FROM cte WHERE rn = 1 UNION SELECT *FROM sku WHERE DAY(price_date) = 1 order by month(date) asc) as a; this one is simpler soln as compared to u
@sourabhpatel38347 ай бұрын
WITH CTE1 AS (SELECT price_date, price, ROW_NUMBER() OVER (PARTITION BY MONTH(PRICE_DATE) ORDER BY PRICE_DATE DESC) RNK, DATEADD(DAY, 1, EOMONTH(price_date)) next_month FROM SKU), CTE2 AS (SELECT price_date, price, price_date AS next_month FROM sku WHERE DAY(price_date) = 1 UNION ALL SELECT price_date, price, next_month FROM CTE1 WHERE RNK = 1) SELECT *, price - LAG(price, 1, price) OVER (ORDER BY next_month) diff FROM CTE2;
@hairavyadav657928 күн бұрын
Nice explanation sir great.....
@harshSingh-if4zb4 ай бұрын
A little mess but getting correct output: select sku_id ,price_date, price, dr, case when month = 0 then price else lg end as final_price from (select *, lag(price,1,0) over(order by price_date) as lg from (select *, dense_rank() over(partition by month1 order by price_date desc) as dr from (select *, concat(left(price_date , 7) , "-01") as month1, datediff(price_date, concat(left(price_date , 7) , "-01")) as month from sku) a)b where dr=1)c;
@amanpratapsingh86317 ай бұрын
Here is my solution in MySQL: with cte as( select *,row_number() over(order by sku_id) as m, year(price_date) as y from sku), cte2 as( select *,concat(y,"-",m,"-","01") as concat_date from cte), cte3 as( select sku_id,price_date,price,STR_TO_DATE(concat_date,"%Y-%m-%d") AS converted_date from cte2 order by converted_date), cte4 as( select *, lag(price) over(order by price) as lag_price, datediff(converted_date,price_date) as dd from cte3) select sku_id,converted_date, case when dd>=0 then price when dd
@srinivasareddybandi9837 ай бұрын
with cte1 as ( select *,row_number() over(partition by sku_id,extract(month from price_date) order by price_date desc) rn from sku ) ,cte2 as ( select sku_id,price_date,price,cast(date_trunc('month',price_date) as date) as dt,lag(price,1,price) over() as lp from cte1 where rn = 1 union all select sku_id,cast(price_date+interval '1 month' as date) as price_date,price, cast(date_trunc('month',price_date+interval '1 month') as date) as dt,price as lp from sku where price_date = (select max(price_date) from sku) ) select dt,new_price,new_price-lag(new_price,1,new_price) over () as diff from ( select dt,price_date, case when dt=price_date then price when dt
@vaibhavverma13407 ай бұрын
Here is my Attempt Sir , Please have a look. with cte as (select *, DATEFROMPARTS(year(price_date),month(price_date),'01')start_of_month ,(case when price_date > DATEFROMPARTS(year(price_date),month(price_date),'01') then lag(price,1) over (partition by sku_id order by price_date) else price end)price_start_of_month from sku) ,eliminate_duplication_months as (select sku_id, start_of_month, price_start_of_month, dense_rank() over (partition by sku_id, start_of_month order by price_date)dr from cte) select sku_id as SKU, start_of_month as [Date] , price_start_of_month as Price, price_start_of_month -lag(price_start_of_month,1,price_start_of_month) over (partition by sku_id order by price_start_of_month)Dif from eliminate_duplication_months where dr =1
@sobermenezes6 ай бұрын
Excellent video Ankit. A query on your second method though: the inner join you used has an incomplete ON clause (on.c.cal_date). How’s that possible?
@atulsharma27896 ай бұрын
Ankit this solution is work or not ?? with cte as( select *, ROW_NUMBER() over(partition by sku_id,year(price_date),month(price_date) order by price_date desc) skudate from sku),cte2 as( select sku_id,price_date,DATEADD(Month,DATEDIFF(Month,1,price_date),0)nextofmonth,price from cte where skudate=1), cte3 as( select sku_id,price_date,price,isnull(lead(price_date) over(order by sku_id),'2023-05-01') nextmonth1 from cte2 ) select sku_id,price_date,price from sku where datepart(day,price_date)=1 union all select sku_id,DATEADD(Month,DATEDIFF(Month,1,nextmonth1),0),price from cte3
@zaravind42937 ай бұрын
Hi Ankit I have one query how to get the alternate characters in upper case remaining in lower case like name is Rahul then output should be RaHuL. how can we achieve this in sql
@ManpreetSingh-tv3rw7 ай бұрын
Could this have been done by the recursive CTE ? Like expanding the rows from Jan 1 to Jan 30 , then feb 1 to feb 29 ? I am trying this approach not sure if it will work. @ankitbansal6
@MubarakAli-qs9qq19 күн бұрын
🎉Sir mast question h
@jay_rana7 ай бұрын
In the 1 st attempt can't we use Union instead of Union All, this will remove the duplicate record with same price date value on 1st day of month cases ??
@vikasvk91747 ай бұрын
Hi Ankit, I have one doubt instead of 2023-01-01 we have 2023-01-10 in that case will not get first recode in our final output ryt ?
@saketarora7 ай бұрын
Doing unions all and then not in?? Could have just done union?
@ankitbansal67 ай бұрын
The price can be different ..
@AmanRaj-uf7wx7 ай бұрын
MYSQL: with cte as ( select *, row_number() over (order by sku_id) as mth, str_to_date((concat(year(price_date), "-", row_number() over (order by sku_id), "-", "01")),"%Y-%m-%d") as updated_date from sku ) ,cte2 as ( select sku_id, price_date, price,updated_date, lag(price,1,0) over (order by price ) as lag_price, datediff(updated_date, price_date) as dd from cte order by updated_date) select sku_id, updated_date, case when dd >= 0 then price when dd
@MITHUNKUMAR-nx8rd7 ай бұрын
Thanks
@Aman-lv2ee7 ай бұрын
We can use union also instead of union all and a subquery: with cte as ( select *,dense_rank()over(partition by sku_id, extract(month from price_date), extract(year from price_date) order by price_date desc) as dk from sku ), cte2 as ( select sku_id, price_date as new_price_date, price from sku where date_part('day', price_date) = 1 union select sku_id, date(date_trunc('month',price_date+INTERVAL '1 month')) as new_price_date , price from cte where dk =1 ) select *, lag(price,1,10)over(order by extract(month from new_price_date)), price-lag(price,1,10)over(order by extract(month from new_price_date)) as difference from cte2
@vinil92122 ай бұрын
smart work handling the NULL, but you could have just used COALESCE
@nash_life7 ай бұрын
Good Explanations Sir. I failed 3 interviews in the past 3 days because of SQL. i am not sure why I am not able to build solutions. I hope to learn from your videos.
@ankitbansal67 ай бұрын
Don't worry keep practicing
@srikarrar2617 ай бұрын
Sir, please Create Calendar Table video
@madhurimadas62607 ай бұрын
I did like this with cte as(SELECT *, DATEADD(month, DATEDIFF(month, 0, price_date) + 1, 0) AS first_day_of_month, row_number()over(partition by year(price_date),month(price_date) order by price_date)as rnk, lead(price) over(partition by year(price_date),month(price_date) order by price_date)as pre FROM sku) select cte.sku_id,case when pre is null then price else pre end as price,first_day_of_month from cte where cte.rnk=1 union all select sku_id,price,price_date from sku where datepart(day,price_date)=1 order by first_day_of_month
@SreemantaKesh6 ай бұрын
great question
@story_teller_Is4 ай бұрын
sir aap tottle hn kya
@ankitbansal64 ай бұрын
Haan m totla hoon. Mere papa bhi totle hain..Mera pura khandaan totla hai. Hum sab TA ko TA bolte hain
@story_teller_Is4 ай бұрын
@@ankitbansal6 😃lagta h aap bhavuk hogye😁
@ankitbansal64 ай бұрын
@@story_teller_Is haha just kidding 😂
@june17you7 ай бұрын
Hi can anyone help me on what is the equivalent function of datetrunc in mysql
@ankitbansal67 ай бұрын
You need to use the extract function
@sammail967 ай бұрын
For oracle sql also extract function works same way as datetrunc@@ankitbansal6
@ManishChauhan-fb7uz7 ай бұрын
select sku_id as SKU ,price_date as Date ,price ,prev_price-lag(prev_price,1,prev_price) over(order by price_date) as Diff from ( select * ,lag(price, 1, price) over(order by date_part('month',price_date)) as prev_price ,rank() over(partition by date_part('month', price_date) order by date_part('day',price_date)) as rnk from sku ) as t where rnk = 1
@sabesanj55097 ай бұрын
WITH cte AS ( SELECT SKU, DATE, PRICE ROW_NUMBER OVER (PARTITION BY SKU ORDER BY DATE) AS rnk FROM prices ) SELECT SKU, DATEFORMAT(DATE, '%y%M-01') AS start_of_month, PRICE FROM cte WHERE rnk = 1;