Famous SQL Interview Question | First Name , Middle Name and Last Name of a Customer

[@ankitbansal6]

Join 100 days of SQL challenges where I have created hand picked SQL interview questions to sky rocket your SQL skills. 100daysofsql.com/

[@neerajpathak7585]

This also seems to work fine for me : SELECT split_part(customer_name,' ',1) as first_name ,case when split_part(customer_name,' ',3) ='' then '' else split_part(customer_name,' ',2) end as second_name ,case when split_part(customer_name,' ',3) ='' then split_part(customer_name,' ',2) else split_part(customer_name,' ',3) end as third_name from customers

[@Riteshkumar-r5o]

Simple and easy solution for PostgreSQL thanks buddy for your effort

[@neerajpathak7585]

@@Riteshkumar-r5o 👍🏻

[@addhyasumitra90]

My approach: bit compilcated but inspired by your method and with diff approach: with CTE as ( select customer_name, len(customer_name) - len(REPLACE(customer_name,' ','')) as spaces, value, ROW_NUMBER() OVER(order by customer_name ) as rn from customers cross apply string_split(customer_name, ' ')), CTE2 as ( select customer_name, case when spaces=0 then customer_name when ROW_NUMBER() over(partition by customer_name order by rn)=1 then value end as first_name ,case when spaces

[@GiriPrasath.D]

From this question, i learned, Char index, substring, left and right functions, you are SQL hero Ankit. and how to use the substring with len and charindex to extract the first , middle and last name.

[@ankitbansal6]

Keep learning 😊

[@architsrivastava6649]

My interview in 30 mins and was watching your Sql playlist ❤️

[@bhumikalalchandani321]

good luck!!

[@ankitbansal6]

All the best 🙂

[@CricketLover-qy9nn]

If u don't mind can you share your interview experience.

[@dfkgjdflkg]

Not surprised endlessly impressive mastery that can only be envied. thanks

[@usharanikallubhavi7466]

Hi Ankit. To extract middle_name, can we write SUBSTRING(customer_name, first_space_position+1, second_space_position-1)

[@pradeeppatil7168]

Hi Ankit, i guess i have made it more simple, just check it. select *, case when len(empname)-len(replace(empname,' ',''))=1 OR len(empname)-len(replace(empname,' ',''))=2 then substring(empname,1,charindex(' ',empname)) ELSE EmpName end as F_Name, case when len(empname)-len(replace(empname,' ',''))=2 then substring(empname,charindex(' ',empname)+1,CHARINDEX(' ',empname,CHARINDEX(' ',empname)+1)-charindex(' ',empname)) end as M_Name, case when len(empname)-len(replace(empname,' ',''))=1 OR len(empname)-len(replace(empname,' ',''))=2 then substring(empname,charindex(' ',empname,5)+1,len(empname)) end as L_Name from Employee

[@AbhaKumari-oc1fj]

My solution: select name, split_part(name, ' ', 1) as first_name, split_part(name, ' ', 2) as scn_name, split_part(name, ' ', 3) third_name from names this working perfectly, please give suggestion or advice if any point i missed.

[@jhonsen9842]

Could you please post more Data Engineering SQL Questions . I think these questions are more alligned to Data Analyst which is cool also. Looking some super hard questions on Join and CTE,Subqueries.

[@ayushmi77al]

In postgresql it's very easy, we can use split_part function.

[@ankitbansal6]

Right but thats not available in most of other databases.

[@ayushmi77al]

So true, your SQL interview question series so very helpful, thank you so much for that 🙏

[@anudipray4492]

Oracle function much better

[@vsagar4b38]

Liked the Logic and Explanation, superb Enjoyed. Thanks Ankit

[@ankitbansal6]

Glad you liked it

[@Ak12345-g]

Thanks for sharing good problems❤ worth watching

[@ankitbansal6]

Glad you enjoyed it

[@shyamu431]

Thanks Ankit, your questions and the way you solve it, is amazing. It's very good way to build the logical understanding by wathcing your videos. Coming to this particular video. If the name has more than 3 words i.e. Vijay Pratap Singh Rathore. The query will become more complex. Is there any other method to solve it. Once again, I appreciate your efforts in making these wonderful tutorials.

[@ankitbansal6]

Yes it will become complex. Also you need to decide what you want to keep in the middle name and last name. In postgres and redshift we have a split part function which can make the solution easy . I hope other databases introduce that function.

[@gauravkakhani7165]

Using right function last name: Case when no_of_spaces= 0 then null when no_of_spaces= 1 then right(customer_name,len(customer_name) - firstspaceposition) else right(customer_name,len(customer_name)- second space position)end as lastname from cte;

[@ankitbansal6]

Perfect 💪

[@rahulkanojiya6256]

another way to solve the same question without using string fuctions : with cte as( select customer_name, value ,row_number() over (partition by customer_name order by customer_name) as rnk ,count(value) over(partition by customer_name order by customer_name) as cnt from customers cross apply string_split(customer_name , ' ') ) ,cte2 as ( select customer_name ,case when rnk = 1 then value end as first_name ,case when rnk = 2 and cnt = 3 then value end as middle_name ,case when (rnk = 2 and cnt = 2) or (rnk = 3 or cnt = 3) then value end as last_name from cte ) select customer_name , max(first_name) as first_name , max(middle_name) as middle_name , max(last_name) as last_name from cte2 group by customer_name

[@srinubathina7191]

Thank you Bro again explanation is next level

[@yakkaluruvijaysankar787]

Your mic quality is not good. There is no clarity on what you are explaining. The information is very good and informative. Please do more videos like this.

[@roshangangurde7188]

Great explanation sir 🙏🙏

[@DE_Pranav]

with cte as( select * from customers cross apply string_split(customer_name,' ') ) ,cte2 as( select *, ROW_NUMBER() over(partition by customer_name order by(select null)) as rn, count(*) over (partition by customer_name) as cnt from cte ) select customer_name, max(case when rn=1 then value end) as firstname, max(case when rn=2 and cnt=3 then value end) as middlename, max(case when (rn=2 and cnt=2) or rn=3 then value end) as lastname from cte2 group by customer_name

[@ss-hm6ey]

Hi Sir, i recently gave an interview at LTIMINDTREE and the very first question they asked was there are three tables and can we use left and right join on them. I didnt understand the question also clearly. Pls make a video on this and explain in detail. Thank you.

[@mixedentertainment_hub7217]

We can

[@Riteshkumar-r5o]

Hi Ankit, Good day! Can you please provide the solution for this question in PostgreSQL, as Position function in Postgre takes only two arguments, that's creating trouble getting the position of second space in the customer_name field. If not possible can you please just suggest me which function to use here for getting second space position in postgresql. Anyone from the community can suggest please.

[@deepeshmatkati3058]

Great explanation Ankit

[@SwathiRavichandran-xh8wq]

Hi Ankit ..your videos are good . Can you help how this can be achieved in oracle sql

[@TanmayModi06]

This Works fine: with cte as( select *, ROW_NUMBER() over(order by customer_name) as rn from customers ), cte1 as( select value, rn, ROW_NUMBER() over(partition by rn order by rn) as rk from cte cross apply string_split(customer_name, ' ') ), cte2 as( select *, case when rk = 1 then value end as firstname, case when rk 1 and rk (select Max(rk) from cte1 where rn = a.rn) then value end as middle_name, case when rk = (select Max(rk) from cte1 where rn = a.rn) and rk 1 then value end as lastname from cte1 a ) select STRING_AGG(firstname,'') as firstname, STRING_AGG(middle_name,'') as middle_name, STRING_AGG(lastname,'') as lastname from cte2 group by rn

[@vikramjitsingh6769]

Those who are looking for MySQL version Solution - select *, substring_index(customer_name, ' ',1) as First , case when x >= 1 then substring_index(customer_name, ' ',-1) else null end as last , case when x >= 2 then substring_index(substring_index(customer_name, ' ',2),' ',-(x-1)) else null end as middle from (select *, length(customer_name) - length(replace(customer_name,' ','')) as x from customers)x

[@jececdept.9548]

can use locate as well

[@monuoriginal7425]

brother electoral bond par ek baar join laga ke bataona problem aarahi hai meko i am beginer also ....because data duplicates.....

[@mangeshbhumkar2075]

In bigquery possible with split function with safe_offset(0),safe_offset(1) and safe_offset(2)

[@007SAMRATROY]

So if there are N number of words in the name, we will have to derive N - 1 number of space positions right?

[@Ashu23200]

Superb!

[@akashsonone2838]

Hello Ankit , I've attempted another approach. Please inform me if it's functioning correctly in every corner case as well. WITH CTC AS( SELECT *, LEN(CUSTOMER_NAME) - LEN(REPLACE(CUSTOMER_NAME,' ','')) AS NO_OF_SPACES, LEFT(CUSTOMER_NAME, CHARINDEX(' ',CUSTOMER_NAME)) AS FIRST_NAME, RIGHT(CUSTOMER_NAME, CHARINDEX(' ',REVERSE(CUSTOMER_NAME))) AS LAST_NAME FROM CUSTOMERS ) SELECT CASE WHEN NO_OF_SPACES = 0 THEN CUSTOMER_NAME ELSE FIRST_NAME END AS FIRST_NAME, CASE WHEN NO_OF_SPACES = 2 THEN SUBSTRING(CUSTOMER_NAME,LEN(FIRST_NAME)+2, LEN(CUSTOMER_NAME)-LEN(FIRST_NAME)-LEN(LAST_NAME)) END AS MIDDLE_NAME, CASE WHEN LEN(LAST_NAME) = 0 THEN NULL ELSE LAST_NAME END AS LAST_NAME FROM CTC

[@Soul-f3v]

with cte as( select *,ROW_NUMBER() over(partition by customer_name order by (select null)) as name_part,COUNT(*) over (partition by customer_name) as total_parts from customer cross apply string_split(customer_name,' ')) select customer_name, max(case when name_part =1 then value else null end) as first_name, max(case when name_part =2 and total_parts > 2 then value else null end) as middle_name, max(case when name_part=3 or (name_part=2 and total_parts =2) then value else null end) as last_name from cte group by customer_name;

[@PraveenSinghRathore-df3td]

Easiest MYSQL solution: select substring_index(customer_name,' ',1) as first_name, case when (length(customer_name) - length(replace(customer_name,' ',''))) = 0 then null else substring_index(substring_index(customer_name,' ',2),' ',-1) end as middle_name, case when (length(customer_name) - length(replace(customer_name,' ',''))) = 0 then null else substring_index(customer_name,' ',-1) end as last_name from customers;

[@NehaAgarwal-l8l]

with CTE as ( select * from customers cross apply string_split(customer_name,' ') ), CTE2 as (select *,count(*) over(partition by customer_name) as words_count,row_number() over(partition by customer_name order by (select null)) as rn from CTE) select customer_name,max(case when words_count in (1,2,3) and rn=1 then value end) as first_name , max(case when words_count in (3) and rn=2 then value end) as middle_name , max(case when words_count in (2) and rn=2 or words_count in (3) and rn=3 then value end) as last_name from CTE2 group by customer_name

[@ishika7585]

with cte as ( select customer_name, value AS part, ROW_NUMBER() over (partition by customer_name order by customer_name) AS part_number from customers cross apply STRING_SPLIT(customer_name, ' ') ) ,final as ( select customer_name, max(case when part_number = 1 then part end) as first_name, max(case when part_number = 2 then part end) as middle_name, max(case when part_number = 3 then part end) as last_name from cte group by customer_name ) select * from final;

[@boogieman8827]

Is Syllabus for Data engineer and Data Analyst the same? How much are the similarities?

[@Tech_world-bq3mw]

its different

[@boogieman8827]

@@Tech_world-bq3mw how much SQL and python is same in Data Engineering vs Data Analytics?

[@SiiitiiFreelancing-jl3ty]

in Postgres? with strpos or position? how is second space found?

[@rajatbhat2074]

Hi Ankit, Tried below solution in PostgreSQL, its working. Let me know your thoughts. with cte as ( select customer_name, length(customer_name)- length(replace(customer_name,' ','')) as no_of_spaces from customers ) select customer_name, case when no_of_spaces >=0 then split_part(customer_name, ' ', 1) end as first_name, case when no_of_spaces =1 then split_part(customer_name, ' ', no_of_spaces+1) end as last_name from cte;

[@ankitbansal6]

It's good but the split part function is not available in most other databases

[@boogieman8827]

how much SQL and python is same in Data Engineering vs Data Analytics?

[@myselfrithish]

What is meant by SQL and t SQL is it necessary for data analytics job

[@arvindgurjar3300]

Thank you so much

[@ankitbansal6]

You're most welcome

[@sarathmaya6083]

@ankitbansal6 please review RIGHT function for last name case when no_of_space=0 then null --when no_of_space=1 then SUBSTRING(customer_name,first_space_position+1,first_space_position) --when no_of_space=2 then SUBSTRING(customer_name,second_space_position+1,second_space_position) when no_of_space=1 then RIGHT(customer_name,first_space_position) when no_of_space=2 then RIGHT(customer_name,second_space_position-first_space_position-no_of_space) end as last_name

[@vinil9212]

can middle name be extracted with LEFT or RIGHT?

[@apurvasaraf5828]

with cte as (select *,LEN(customer_name)-len(REPLACE(customer_name,' ','')) as l ,CHARINDEX(' ',customer_name) as f, CHARINDEX(' ',customer_name,CHARINDEX(' ',customer_name)+1) as s from Customers) select *,case when l=0 then customer_name else substring(customer_name,1,f-1) end as firstn, case when l

[@adharshsunny5154]

Please do create AWS videos

[@NextGen_Tech_Hindi]

Same

[@prakritigupta3477]

This is the solution in PostgresSQL select split_part(customer_name,' ',1) as first_name, split_part(customer_name,' ',2) as middle_name, split_part(customer_name,' ',3) as last_name from customers;

[@ankitbansal6]

If there is no middle name then your query will give last name as null and middle name will be last name.

[@DEwithDhairy]

PySpark Version of this problem : kzbin.info/www/bejne/kKOZhoucqdOkhbc

[@paritoshjoshi5623]

with temp as( select customer_name, length(customer_name)-length(replace(customer_name,' ','')) ct from customers) select substring_index(customer_name,' ',1) as first_name ,if(ct=2,substring_index(substring_index(customer_name,' ',-2),' ',1) ,null) as middle_name ,if(ct=1 or ct=2,substring_index(customer_name,' ',-1),null) as last_name from temp;

[@akashmukherjee9011]

with cte as ( select customer_name, length(customer_name)-length(replace(customer_name,' ','')) as no_of_spaces from customers) select *, case when no_of_spaces = 0 then customer_name when no_of_spaces = 1 then substring_index(customer_name,' ',1) when no_of_spaces = 2 then substring_index(customer_name,' ',1) else null end as first_name, case when no_of_spaces = 2 then substring_index(substring_index(customer_name,' ',-2),' ',1) else null end as middle_name, case when no_of_spaces = 1 then substring_index(customer_name,' ',-1) when no_of_spaces = 2 then substring_index(customer_name,' ',-1) else null end as last_name from cte

[@chrishkumar1250]

MYSQL SELECT substring_index(customer_name," ",1) as first_name, if (length(customer_name) - length(replace(customer_name, ' ', '')) > 1 , substring_index(substring_index(customer_name," ",2), " ",-1), Null) as second_name, if (length(customer_name) - length(replace(customer_name, ' ', '')) >= 2 , substring_index(substring_index(customer_name," ",3), " ",-1), Null) as third_name from customers

[@NextGen_Tech_Hindi]

Love you 💓

[@naveenvjdandhrudu5141]

Bro just post some easy interview questions in sql

[@ankitbansal6]

Ok next time

[@kedarwalavalkar6861]

My solution : with cte as ( select *, substring_index(customer_name,' ',1) as a ,substring_index(substring_index(customer_name,' ',2),' ',-1) as b ,substring_index(customer_name,' ',-1) as c ,round((length(customer_name) - length(replace(customer_name,' ','')))/length(' '),0) as leng from custs ) select a as first_name ,case when leng = 2 then b else null end as middle_name ,case when leng = 1 then b when leng = 2 then c else null end as last_name from cte;

[@anirbanbiswas7624]

SOLUTION FOR MYSQL BY USING SPACE_COUNT ONLY with cte as(select *, length(customer_name)-length(replace(customer_name,' ','')) as space_count from customers1) select customer_name, case when space_count=0 or space_count=1 or space_count=2 then substring_index(customer_name,' ',1) end as trial, case when space_count=2 then substring_index(substring_index(customer_name,' ',2),' ',-1) else null end as mid_name, case when space_count=2 or space_count=1 then substring_index(customer_name,' ',-1) else null end as last_name from cte

[@sahilummat8555]

;with cte as ( select *, LEN(customer_name)- len(REPLACE(customer_name,' ','')) as spaces, CHARINDEX(' ',customer_name) as space_position, CHARINDEX(' ',customer_name,CHARINDEX(' ',customer_name)+1) as space_position_2 from customers) select *, case when spaces=0 then customer_name when spaces!=0 then left(customer_name,space_position-1) end as first_name , case when spaces>1 then SUBSTRING(customer_name,space_position+1,space_position_2-space_position) end as middle_name , case when spaces=1 then right(customer_name,len(customer_name)-space_position) when spaces>1 then right(customer_name,len(customer_name)-space_position_2) end as last_name from cte

[@DIVYATRIPATHI-w9f]

Hello Sir in my sql it is showing error. I have used INSTR Function

[@ankitbansal6]

Cool

[@skkholiya]

with cte as( select if(instr(customer_name," ")>0, substring(customer_name,1, instr(customer_name," ")),customer_name) first_name, trim(substr(customer_name,instr(customer_name," "), length(customer_name))) remaining_name from customers ) select first_name, if(instr(remaining_name," ")>0, substring(remaining_name,1,instr(remaining_name," ")),"") middle_name, if(instr(remaining_name," ")>0, substring(remaining_name,instr(remaining_name," "), length(remaining_name)),remaining_name) last_name from cte;

[@naveenvjdandhrudu5141]

There is only medium & complex in your play list

[@priyanshushak7705]

*for mysql versions which does not have charindex and its equivalent or if there are multiple middle names* : use below approach with cte_spaces as ( select * , length(customer_name)- length(replace( customer_name, ' ','')) as no_of_spaces from customers ) select * , substring_index(customer_name, " ",1) as first_name, case when no_of_spaces > 1 then substring_index(substring_index(customer_name, " ", (-1 * no_of_spaces)), " ", no_of_spaces -1) end as middle_name, case when no_of_spaces > 0 then substring_index(customer_name, " ",-1) end as last_name from cte_spaces

[@zainaltaf4935]

Just confused from where u find such questions 😅

[@saipranay1318]

I feel the same !! lol 😀

[@anudipray4492]

Anyone from oracle can do it

[@hsk7715]

it's look difficult

[@Tech_world-bq3mw]

You logic will fail if there is space in starting or in ending of string.

[@ankitbansal6]

In that case you can just trim the customer name in first cte and then as it is it will work

[@Hope-xb5jv]

Logic fail if two or three space between name

[@ankitbansal6]

You can trim to single space first

[@AmanRaj-p8w]

MySQL Solution: with cte as ( select customer_name, (length(customer_name) - length(replace(customer_name, ' ', '')) + 1) as cnt_of_words from customers ) SELECT CASE WHEN cnt_of_words = 1 THEN customer_name -- Only one word, consider it as the first name WHEN cnt_of_words = 2 THEN SUBSTRING_INDEX(customer_name, ' ', 1) -- Two words, consider the first word as first name WHEN cnt_of_words = 3 THEN SUBSTRING_INDEX(customer_name, ' ', 1) -- Three words, consider the first word as first name END AS first_name, CASE WHEN cnt_of_words = 3 THEN SUBSTRING_INDEX(SUBSTRING_INDEX(customer_name, ' ', 2), ' ', -1) -- Three words, consider the second word as middle name END AS middle_name, CASE WHEN cnt_of_words >= 2 THEN SUBSTRING_INDEX(customer_name, ' ', -1) -- At least two words, consider the last word as last name END AS last_name FROM cte;

[@rubyshorts281]

for mysql users WITH cte AS ( SELECT *, LENGTH(customer_name) - LENGTH(REPLACE(customer_name, ' ', '')) AS spaces FROM customers ) SELECT CASE WHEN spaces = 2 THEN SUBSTRING_INDEX( SUBSTRING_INDEX(customer_name, ' ', -spaces), ' ', spaces - 1 ) ELSE NULL END AS middle, CASE WHEN spaces >= 1 THEN SUBSTRING_INDEX(customer_name, ' ', -1) -- Last name ELSE NULL END AS last FROM cte;

[@snehalpattewar7864]

SELECT SUBSTRING_INDEX(name, ' ', 1) AS first_name, CASE WHEN LENGTH(name) - LENGTH(REPLACE(name, ' ', '')) > 1 THEN SUBSTRING_INDEX(SUBSTRING_INDEX(name, ' ', -2), ' ', 1) ELSE NULL END AS middle_name, SUBSTRING_INDEX(name, ' ', -1) AS last_name FROM your_table_name;

[@samiphani2473]

SELECT customer_name, CASE WHEN size(split(customer_name, ' ')) = 2 THEN split(customer_name, ' ')[0] ELSE split(customer_name, ' ')[0] END AS First_Name, CASE WHEN size(split(customer_name, ' ')) = 2 THEN NULL ELSE split(customer_name, ' ')[1] END AS Middle_Name, CASE WHEN size(split(customer_name, ' ')) = 2 THEN split(customer_name, ' ')[1] ELSE split(customer_name, ' ')[2] END AS Last_Name FROM customers;

[@iramansari3625]

what if we have more then 2 space then @ankitbansal6 ?

[@nipunshetty9640]

Hii Ankit bansal from my Side small request, As told by you in UR LINKEDIN PROFILE, Supply chain Analytics, Could you make one Video of SUPPLY CHAIN ANALYTICS BY TAKING PROCUREMENT SUPPLY CHAIN DATA SET AND Do DATA ANALYSIS so it will help me and even every audience please My Request Sir @ankitbansal6

[@krsnabnsl]

With PySpark DataFrame API df = customers_df.withColumn("splitter", split(customers_df["customer_name"], " ")) df1 = df.withColumn("len", size("splitter")) df2 = df1.withColumn("FirstName", col("splitter")[0])\ .withColumn("MiddleName", when(col("len")>2, col("splitter")[1]))\ .withColumn("LastName", when(col("len")==2, col("splitter")[1]).otherwise(col("splitter")[2]))\ .drop("splitter", "len") df2.show()