SQL Test Based on Real Interview | SQL Interview Questions and Answers
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- Опубликовано: 26 июл 2024
- In this video we will solve a complete SQL test consist on 7 interview questions. This would be a great exercise to practice SQL.
Start your data analytics journey: www.namastesql.com/
script:
CREATE TABLE users (
USER_ID INT PRIMARY KEY,
USER_NAME VARCHAR(20) NOT NULL,
USER_STATUS VARCHAR(20) NOT NULL
);
CREATE TABLE logins (
USER_ID INT,
LOGIN_TIMESTAMP DATETIME NOT NULL,
SESSION_ID INT PRIMARY KEY,
SESSION_SCORE INT,
FOREIGN KEY (USER_ID) REFERENCES USERS(USER_ID)
);
-- Users Table
INSERT INTO USERS VALUES (1, 'Alice', 'Active');
INSERT INTO USERS VALUES (2, 'Bob', 'Inactive');
INSERT INTO USERS VALUES (3, 'Charlie', 'Active');
INSERT INTO USERS VALUES (4, 'David', 'Active');
INSERT INTO USERS VALUES (5, 'Eve', 'Inactive');
INSERT INTO USERS VALUES (6, 'Frank', 'Active');
INSERT INTO USERS VALUES (7, 'Grace', 'Inactive');
INSERT INTO USERS VALUES (8, 'Heidi', 'Active');
INSERT INTO USERS VALUES (9, 'Ivan', 'Inactive');
INSERT INTO USERS VALUES (10, 'Judy', 'Active');
-- Logins Table
INSERT INTO LOGINS VALUES (1, '2023-07-15 09:30:00', 1001, 85);
INSERT INTO LOGINS VALUES (2, '2023-07-22 10:00:00', 1002, 90);
INSERT INTO LOGINS VALUES (3, '2023-08-10 11:15:00', 1003, 75);
INSERT INTO LOGINS VALUES (4, '2023-08-20 14:00:00', 1004, 88);
INSERT INTO LOGINS VALUES (5, '2023-09-05 16:45:00', 1005, 82);
INSERT INTO LOGINS VALUES (6, '2023-10-12 08:30:00', 1006, 77);
INSERT INTO LOGINS VALUES (7, '2023-11-18 09:00:00', 1007, 81);
INSERT INTO LOGINS VALUES (8, '2023-12-01 10:30:00', 1008, 84);
INSERT INTO LOGINS VALUES (9, '2023-12-15 13:15:00', 1009, 79);
-- 2024 Q1
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (1, '2024-01-10 07:45:00', 1011, 86);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (2, '2024-01-25 09:30:00', 1012, 89);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (3, '2024-02-05 11:00:00', 1013, 78);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (4, '2024-03-01 14:30:00', 1014, 91);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (5, '2024-03-15 16:00:00', 1015, 83);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (6, '2024-04-12 08:00:00', 1016, 80);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (7, '2024-05-18 09:15:00', 1017, 82);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (8, '2024-05-28 10:45:00', 1018, 87);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (9, '2024-06-15 13:30:00', 1019, 76);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (10, '2024-06-25 15:00:00', 1010, 92);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (10, '2024-06-26 15:45:00', 1020, 93);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (10, '2024-06-27 15:00:00', 1021, 92);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (10, '2024-06-28 15:45:00', 1022, 93);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (1, '2024-01-10 07:45:00', 1101, 86);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (3, '2024-01-25 09:30:00', 1102, 89);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (5, '2024-01-15 11:00:00', 1103, 78);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (2, '2023-11-10 07:45:00', 1201, 82);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (4, '2023-11-25 09:30:00', 1202, 84);
INSERT INTO LOGINS (USER_ID, LOGIN_TIMESTAMP, SESSION_ID, SESSION_SCORE) VALUES (6, '2023-11-15 11:00:00', 1203, 80);
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It takes a lot of effort to make these videos. Please do hit the like button and subscribe to the channel.
anything for you majesty
Ankit I do like and watch your videos regularly, but you also know where the crowd goes and which kind of channels got millions of subscribers.🙂
@@sushantkumar7450 please eloborate 😊
@@ankitbansal6 ankit can you make an class on complete de i will for sure subscribe it
@@ankitbansal6
Hi Ankit,
I was not commenting on the quality of your videos. for me, your playlist is most valuable SQL vedios I have ever gone through in youtube.
I just wrote that comment on a lighter note regarding people watching mostly useless family vlogs and some other channels and those creators get subscribes in millions. It is a general trend in all types of youtube channels let it be politics, spiritual, sports.😊
I wish more folks watch your channel before giving interviews , solid stuff. Infact , they should pay you 10% post getting a new job ;)
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)
Thanks Ankit for providing these interview questions
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
You are mind-blowing
Really appreciate your efforts.. always happy to learn from you..
Your Lecture is awesome sir. 🙏🙏
Brilliantly Explained sir .... Mindblowing ... 🙂
one of the best video and as always nice explanation Ankit
Thank you @Ankit. Please share the questions as you did the CREATE & INSERT statements. Thank you once again.
Learned many new date functions, from this video, do it more
Good one !!
AWSM Sir!!
Thank you Ankit, it's really interesting and cool stuff
Thank you🙏
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..
Well done
Hi Ankit,
I really appreciate your efforts. I learnt a lot from your videos and I've become confident in SQL
Thanks a ton.
It's my pleasure
very good session sir
Hi Ankith,
It's one of the best SQL videos, and you have amazing explanation skill.
Glad you liked it!
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...
❤❤❤❤❤❤❤❤ i don't have any words 🎉🎉🎉🎉🎉🎉
Love you 🤟 bro
awesome and straight to the point.. rocked it :)
Cheers 🍻
Superb video Ankit
Thank you so much 😀
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!
In mysql :
select count(distinct USER_ID) user_count, count(SESSION_ID) as session_count,
min(LOGIN_TIMESTAMP) as min_date,
date_sub(min(LOGIN_TIMESTAMP), interval dayofmonth(min(LOGIN_TIMESTAMP)) -1 day) as first_day
from LOGINS
group by QUARTER(LOGIN_TIMESTAMP)
could you share any other approach I am happy to learn if there is any better or optimized approach
-- 2. Quarter Analysis [Oracle SQL]
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;
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??
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
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."
I think I missed sorting it 😊 thanks for pointing out 🙏
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.
Great video Ankit... nothing major, I think in question 5 it should be max(score) not sum
We need to do sum at user and day level first then find the max for each day using row number
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)
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)
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
Check this
Complex SQL Questions for Interview Preparation: ruclips.net/p/PLBTZqjSKn0IeKBQDjLmzisazhqQy4iGkb
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.....
These questions asked for a experienced role or 1-2 year experience?
3+ years of experience
Hello ankit, In question number 5 you missed user_name
i was getting this error with datetrunc function '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.
Share data and expected output
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
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.....
Install SQL server or MySQL
@@ankitbansal6 sir please suggest me any video because I am facing issue while installing it was showing monthly charges like that 🙏🙏🙏🙏
@@ankitbansal6 sir could you provide me link I am using windows and one more thing sir please provide me guidelines sir for data analytics if u help me out it will be helpful for me 🙏🙏
Why DATETRUNC is not working in my SQL server 😢 it says 'datetrunc' is not a recognized built-in function name.
Maybe older version
@@dasoumya I guess DATETRUNC was introduced as a party of SQL server 2022.
You might wanna go through Microsoft's documentation to see if your version of SQL server supports this or not
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
It will work as we are doing the date trunc of quarter not month
@@ankitbansal6 yes u r right
@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
@ankitbansal please reply on this..I have an interview tomorrow...
@ankitbansal I just want to know what will be the output when there is blank row in one table
@@ritudahiya6223
Blank space will be considered as a another character, it is not same as null. Two nulls aren't equal, blanks are same as 1=1,2=2.
Thanks for this question. I created a table and figured out myself. Adding details below, so that you can also try.
CREATE TABLE T (name char(20));
insert into t values
('a'),
('b'),
('c'),
('a'),
('a'),
(''),
('b'),
(null);
select * from T;
CREATE TABLE T1 (ch char(20));
insert into t1 values
('b'),
('a'),
(''),
('c'),
('a'),
(''),
('d'),
(null);
left join : 12 rows
right join : 13 rows
inner join : 11 rows
Inner - 2
Left - 6
Right - 5
Yes inner join will be 2 and not 11 rows..anyways thanks for replying
I am waiting for India to win then I will get 50% discount 😋
Use code INDIA
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;
Great work 👏
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';