SQL Test Based on Real Interview | SQL Interview Questions and Answers 

It takes a lot of effort to make these videos. Please do hit the like button and subscribe to the channel.

Hi Ankit, I appreciate your learning session. Could you please make some videos apart from SQL queries like SQL Dashboard operation, basic options, error handling part, trigger, procedure etc...

thank you for providing

Hi Ankit, You're really awesome, helping so many. Last Question in PostgreSQL: with cte as( select generate_series(min(login_timestamp),max(login_timestamp),'1 day')::date as login_timestamp from logins) select login_timestamp from cte where login_timestamp not in (select distinct login_timestamp::date from logins)

Really appreciate your efforts.. always happy to learn from you..

Learned many new date functions, from this video, do it more

Your Lecture is awesome sir. 🙏🙏

I wish more folks watch your channel before giving interviews , solid stuff. Infact , they should pay you 10% post getting a new job ;)

QUES 6 WAS AWESOME

one of the best video and as always nice explanation Ankit

Great video Ankit... nothing major, I think in question 5 it should be max(score) not sum

very good session sir

Query for Q3)select distinct USER_ID from logins where LOGIN_TIMESTAMP between '2024-01-01' and '2024-01-31' except select distinct USER_ID from logins where LOGIN_TIMESTAMP between '2023-11-01' and '2023-11-30'

26:53 WITH cte AS ( SELECT *, CAST(DAY(login_timestamp) AS SIGNED) - cast((ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY login_timestamp)) as signed) diff FROM logins ) select user_id from cte group by user_id having count(distinct diff) = 1;

32:37 with recursive cte as( select min(date(login_timestamp)) as dt from logins union all select date_add(dt,interval 1 day) from cte where dt < (select max(date(login_timestamp)) from logins)) select c.dt from cte c left join logins l on c.dt = date(l.login_timestamp) where LOGIN_TIMESTAMP is null;

32:45 with recursive cte as( select min(date(login_timestamp)) as dt from logins union all select date_add(dt,interval 1 day) from cte where dt

Question 6: Can we do this? with test as( select user_id, min(LOGIN_TIMESTAMP) as first_day, max(LOGIN_TIMESTAMP) as last_day, count(LOGIN_TIMESTAMP) as ct from logins group by user_id --order by 1 ) select user_id from test where ct = DATEDIFF(day,first_day,last_day)+1

last answer in my sql select * from (WITH RECURSIVE DateRange AS ( SELECT (select min(cast(login_timestamp as date)) as login_date from logins) AS Start_date UNION ALL SELECT DATE_ADD(Start_date, INTERVAL 1 DAY) FROM DateRange WHERE Start_date < '2024-06-28' ) SELECT Start_date FROM DateRange) x where Start_date not in (select distinct cast(login_timestamp as date) as login_date from logins)

Thank you🙏

3rd with cte as (select * from logins where login_timestamp between "2024-01-01" and "2024-01-31" ), ct as (select * from logins where login_timestamp between "2023-11-01" and "2023-11-30") select * from cte where user_id not in (select user_id from ct)

I think this i great but i do think its more advance than most will ask. I cant see many companies asking these questions to juniors. One because they are pretty complex and two because of time.

Superb video Ankit

In the first question There are two tables we can join the two tables and implement the query right? With cte (Select * from logins join users on users.userid= logins.user_id )as l Select user_id,login_timestamp From l As follows in the video..........

I want to learn very complex problems i solved all the questions without any problem as i work on sql daily but i want to improve my problem solving. Kindly guide sir

Hi Ankit Sir, Thank you for this wonderful Case Study.. It took 70 mins for me to solve all the questions.. only for 6th question I used a bit complex query when compared to yours. other wise everthing were good. enjoyed a lot solving these problem statement.. and for last one I used recursive cte..

Hello ankit, In question number 5 you missed user_name

with cte as ( select *,lag(login_timestamp, 1, login_timestamp) over(partition by user_id order by login_timestamp) as l, DATEDIFF(day, lag(login_timestamp, 1, login_timestamp-1) over(partition by user_id order by login_timestamp), login_timestamp) as diff from logins --order by user_id, login_timestamp ) select user_id from cte group by user_id having count(user_id) = sum(diff)

Hi Sir, Thanks a lot for making video and helping student , I understand each and every things but i facing some problems because i am using mysql for q2 i have no option for date_trun . Can you help me in this

If possible, can we use min(user_id), and convert it in simple query: *Twist is, it will give min(user_id) but max(score). select cast(LOGIN_TIMESTAMP as date), max(SESSION_SCORE) max_score, min(user_id) user_id from logins group by cast(LOGIN_TIMESTAMP as date)

Query for 1 ques. not working in mysql. I used date_add funtion compatible with mySQL but it is throwing an error. select user_id , max(LOGIN_TIMESTAMP) from logins group by user_id having max(LOGIN_TIMESTAMP) < date_add(LOGIN_TIMESTAMP,INTERVAL -5 MONTH) Error:- Unknown column in having clause.....

last second in mysql select * from (select USER_ID, min(cast(LOGIN_TIMESTAMP as date)) as first_login, datediff("2024-06-29",min(cast(LOGIN_TIMESTAMP as date))) as no_of_login_required, count(distinct cast(LOGIN_TIMESTAMP as date)) as login_by_user from logins group by user_id order by user_id ) a where no_of_login_required = login_by_user

great explanation

Hi Ankit, it is really helpful, could you please start creating content on AWS and pyspark. Also just to add on in 1st question we need to use user tables and there might be the case user has never logged in so that entry might not come in logins table??

i was getting this error with datetrunc function 'DATETRUNC' is not a recognized built-in function name.'

Hi Ankit Sir You're really awesome, its very helpful added to it if you can add on these question in an seperate file it will be very useful

Hello everyone i got simple solution for question number 6 . please check it select user_id from logins where cast(login_timestamp as date)>= all( select distinct cast(login_timestamp as date) as dt from logins) Thanks,

Why +1 in Q6

How to solve Q7. in MYSQL? Anyone please assist if you know!! Thanks

if i started learning for now than there are very less data engineering internships(if one has data internship it is easy to get a job ,and that person also get hand on experience ) ,,,,so should i do data analyst first only for internship and after that i will study for data engineering for job because it is also a good way , and i need to do an internship because it is in our college criteria , they do not support much if you do not have done internship?? informative video by the way

This video is pretty simple to understand and this expertise will come with practice.. I hope all the folks practice this at least once a day.. thanks so much Ankit

Thank you Ankit, it's really interesting and cool stuff

This is truly amazing service to the analytics community. Thank you brother

Thank you @Ankit. Please share the questions as you did the CREATE & INSERT statements. Thank you once again.

❤❤❤❤❤❤❤❤ i don't have any words 🎉🎉🎉🎉🎉🎉 Love you 🤟 bro

Brilliantly Explained sir .... Mindblowing ... 🙂

Thanks a lot bro! Great content, helped me crack my SQL interview today!!!

Hi Ankit, I really appreciate your efforts. I learnt a lot from your videos and I've become confident in SQL Thanks a ton.

Thanks Ankit for providing these interview questions

Thanks Ankit! Not correcting you at all, but the 2nd question condition is to sort from newest to oldest. Result should be order by datetrunc(quarter,login_timestamp) desc "Nevertheless, exceptional learning as always."

QUESTION 1. SELECT USER_NAME FROM users_new WHERE user_id NOT IN (SELECT USER_ID FROM (SELECT USER_ID,LOGIN_TIMESTAMP, DATEDIFF(MONTH, LOGIN_TIMESTAMP, MAX(LOGIN_TIMESTAMP)over()) as diff, MAX(LOGIN_TIMESTAMP)over() as highest FROM logins) a WHERE diff < 5 )

Hi Ankith, It's one of the best SQL videos, and you have amazing explanation skill.

You are mind-blowing

awesome and straight to the point.. rocked it :)

Why DATETRUNC is not working in my SQL server 😢 it says 'datetrunc' is not a recognized built-in function name.

I have query base city like, i need to check piiza send across city like bangalore, mumbai , delhi etc, pls help how write this code using slq table called pizza.

Just the video I needed to revise! Thanks a ton Ankit!! Just a twist to your 2nd Q, solve the same question (#2) WITHOUT using DATETRUNC function it's quite interesting! Anyone reading this comment, comment your approach in above case, its fun!

for second question : if there is no login in april month (2024-04) then in ur case minimun date will come as (2024-05), then this is not first day of quarter. so your answer might fail in that case , for accuracy below case statement can be used which wont fail in above case : case when qtr=1 then concat(yr,'-01-01') when qtr=3 then concat(yr,'-06-01') when qtr=2 then concat(yr,'-03-01') when qtr=4 then concat(yr,'-09-01') end as day_qtr

Good morning sir, Please help me for setup the sql environment, i show lot of video but some time network adopter error is showing Please suggest me where should i pratice sql or which platform is good for pratice 🙏🙏🙏🙏🙏 i installed oracle 19c.....

These questions asked for a experienced role or 1-2 year experience?

Hi Ankit I have one concern in your solution, you returned those users which are not logged in last 5 months and at having at least once logged in record in logins table, but if user is created but not logged in once in whole history then that user will not be returned by this query..If I am wrong somewhere please correct me.. @ankitbansal6

QUESTION 2: WITH cte AS (SELECT *, MONTH(login_timestamp) AS month, year(login_timestamp) as yr, CASE WHEN MONTH(login_timestamp) < 4 THEN 'Q1' WHEN MONTH(login_timestamp) < 7 AND MONTH(login_timestamp) > 3 THEN 'Q2' WHEN MONTH(login_timestamp) < 10 AND MONTH(login_timestamp) > 6 THEN 'Q3' WHEN MONTH(login_timestamp) 9 THEN 'Q4' END AS QTR FROM logins) SELECT MIN(login_timestamp) AS FIRST_DAY_Q, COUNT(DISTINCT user_id) as user_cnt, COUNT(SESSION_ID) as session_cnt FROM cte GROUP BY QTR

I am waiting for India to win then I will get 50% discount 😋

QUESTION 7: WITH RECURSIVE_CTE AS (SELECT MIN(login_timestamp) AS first_dt FROM logins UNION ALL SELECT first_dt + 1 as first_dt FROM RECURSIVE_CTE ) SELECT CAST(first_dt AS DATE) AS DATE FROM RECURSIVE_CTE WHERE first_dt < (SELECT (MAX(login_timestamp)) FROM logins) AND CAST(first_dt AS DATE) NOT IN (SELECT CAST(login_timestamp AS DATE) FROM logins ) OPTION (MAXRECURSION 1000);

Good one !!

AWSM Sir!!

@Ankit Bansal Can you please write the answer for this query.it will take 2 mins A. B 1. 2 2. 2 1. 3 No value 4 Null. Null There are 2 tables .table A have one row where we have blank value(row 4) .please give output for innere join and left join Eagerly waiting for your reply

Hi,Ankit iam preparing SQL interview questions and Practice Questions with answers but i want in MYSQL if you make upcoming vedios in MYSQL workbench it will be helpful for me its my kind request ..

Hi Ankit! Thanks for the content My approach from q4 onwards 4)WITH quarter_cnt AS( select YEAR(cast(login_timestamp AS Date)) AS login_year,DATEPART(QUARTER,cast(login_timestamp AS Date)) AS login_quarter, COUNT(DISTINCT USER_ID) AS user_cnt, COUNT(DISTINCT SESSION_ID) AS session_cnt from logins GROUP BY YEAR(cast(login_timestamp AS Date)),DATEPART(QUARTER,cast(login_timestamp AS Date))) SELECT CASE WHEN login_quarter=1 THEN CONCAT('01-01-',login_year) WHEN login_quarter=2 THEN CONCAT('01-04-',login_year) WHEN login_quarter=3 THEN CONCAT('01-07-',login_year) WHEN login_quarter=4 THEN CONCAT('01-10-',login_year) END AS qrtr_str_date,session_cnt, (session_cnt-(lag(session_cnt)OVER(ORDER BY login_year,login_quarter)))*100.0/lag(session_cnt)OVER(ORDER BY login_year,login_quarter), user_cnt FROM quarter_cnt --5) WITH score_day AS( SELECT cast(login_timestamp AS Date) AS login_day,user_id, sum(session_score) AS total_score FROM logins GROUP BY cast(login_timestamp AS Date),user_id), rn_cte AS( SELECT *,DENSE_RANK()over(PARTITION BY login_day ORDER BY total_score DESC) AS rn FROM score_day) SELECT * FROM rn_cte WHERE rn=1 ----6) WITH user_login_history AS( SELECT USER_ID, min(cast(LOGIN_TIMESTAMP AS DATE)) as FIRST_LOGIN, MAX(CAST('2024-06-28' AS Date)) AS Last_Date, COUNT(distinct cast(LOGIN_TIMESTAMP AS DATE)) as no_logins FROM logins GROUP BY USER_ID) SELECT *,datediff(DAY,FIRST_LOGIN,Last_Date)+1 AS logins_btw_str_end FROM user_login_history WHERE (datediff(DAY,FIRST_LOGIN,Last_Date)+1)=no_logins ---7) WITH master_dates as( select cast('2023-07-15' AS Date) AS login_date UNION all SELECT DATEADD(day,1,login_date) FROM master_dates WHERE login_date

Tera itna dimag kaise chalta hai,,, tum intelligence ho iska mtlab hum nbi bhai itna chatgpt se answer nikal ke krne walw log hai hum

SELECT * FROM users WHERE last_login < NOW() - INTERVAL '3 months';

Perfect timing! I have an interview in two days. I've solved these in Oracle. -- 1. Usernames that did not login in the past 5 months SELECT user_id, user_name FROM users WHERE user_id NOT IN ( SELECT user_id FROM logins WHERE login_timestamp > ADD_MONTHS(SYSDATE, -5) ); -- 2. Quarter Analysis SELECT TO_CHAR(ADD_MONTHS(TO_DATE('01/01/2024', 'MM/DD/YYYY'), 3 * (TO_CHAR(LOGIN_TIMESTAMP, 'Q') - 1)), 'MM/DD') || '/' || TO_CHAR(MIN(LOGIN_TIMESTAMP), 'YYYY') AS quarter, COUNT(*) AS sessions, COUNT(DISTINCT user_id) AS unique_users FROM logins GROUP BY TO_CHAR(LOGIN_TIMESTAMP, 'Q') ORDER BY 1; -- 3. Users in Jan2024 but not in Nov2023 SELECT DISTINCT user_id FROM logins WHERE TO_CHAR(LOGIN_TIMESTAMP, 'MM/YYYY') = '01/2024' AND user_id NOT IN ( SELECT user_id FROM logins WHERE TO_CHAR(LOGIN_TIMESTAMP, 'MM/YYYY') = '11/2023' ); -- 4. Percent increase from ans2 WITH raw_data AS ( SELECT TO_DATE( TO_CHAR( ADD_MONTHS(TO_DATE('01/01/2024', 'MM/DD/YYYY'), 3 * (TO_CHAR(LOGIN_TIMESTAMP, 'Q') - 1)), 'MM/DD') || '/' || TO_CHAR(MIN(LOGIN_TIMESTAMP), 'YYYY'), 'MM/DD/YYYY') AS quarter, COUNT(*) AS sessions, COUNT(DISTINCT user_id) AS unique_users FROM logins GROUP BY TO_CHAR(LOGIN_TIMESTAMP, 'Q') ORDER BY quarter ) SELECT QUARTER, SESSIONS, unique_users, ROUND((SESSIONS / LAG(SESSIONS, 1, SESSIONS) OVER (ORDER BY quarter)) * 100 - 100) AS pct FROM raw_data; -- 5. Users with highest scores [will show multiple entries is more than one max] WITH cte AS ( SELECT user_id, TRUNC(login_timestamp) AS login_time, SUM(session_score) AS total_score FROM logins GROUP BY user_id, TRUNC(login_timestamp) ) SELECT user_id, login_time, MAX(total_score) AS score FROM cte GROUP BY user_id, login_time ORDER BY user_id, login_time; -- 6. Users logged in everyday WITH cte AS ( SELECT user_id, MAX(trunc(login_timestamp)) OVER (PARTITION BY user_id) AS max_date, MIN(trunc(login_timestamp)) OVER (PARTITION BY user_id) AS min_date, COUNT(DISTINCT trunc(login_timestamp)) OVER (PARTITION BY user_id) AS distinct_dates_count FROM logins ) SELECT DISTINCT user_id FROM cte WHERE max_date - min_date = distinct_dates_count - 1; -- 7. Dates without login WITH cte (start_date) AS ( SELECT MIN(TRUNC(login_timestamp)) AS start_date FROM logins UNION ALL SELECT start_date + 1 FROM cte WHERE start_date < (SELECT MAX(TRUNC(login_timestamp)) FROM logins) ) SELECT * FROM cte WHERE start_date NOT IN ( SELECT DISTINCT TRUNC(login_timestamp) FROM logins ) ORDER BY 1;