SQL Interview Questions

Yashasvi V. - "WITH RECURSIVE fibonacci_series AS ( SELECT 1 AS n, 0 AS fib1, 1 AS fib2 UNION ALL SELECT n + 1 AS n, fib2 AS fib1, fib1 + fib2 AS fib2 FROM fibonacci_series WHERE n < 20 -- Limit the series to 20 numbers ) SELECT n, fib1 AS fib FROM fibonacci_series ORDER BY n; `"See full answer

Peter W. - "In the question it says: "above the overall average total posts", which to me implying a >, yet in the solution it uses >= Caused me 1 hr to find out. plz fix"See full answer

Christian B. - "with cte as (select ts.employee_id, e.name, t.id as test_id, max(distinct ts.score) as total from test_results as ts join tests as t on ts.test_id = t.id join employees as e on ts.employee_id = e.id group by ts.employee_id, e.name, t.id) select employee_id, name as employee_name, sum(total) as total_score from cte group by employee_id, employee_name order by total_score desc, employee_id asc ;"See full answer

Erjan G. - "1) select avg(session) from table where session> 180 2) select round(sessiontime/300)*300 as sessionbin, count() as sessioncount from table group by round(sessiontime/300)300 order by session_bin 3) SELECT t1.country AS country_a, t2.country AS country_b FROM ( SELECT country, COUNT(*) AS session_count FROM yourtablename GROUP BY country ) AS t1 JOIN ( SELECT country, COUNT(*) AS session_count FROM yourtablename `GROUP BY countr"See full answer

Zacharias E. - "SELECT pro.id, pro.title, pro.budget, COUNT(employeeid) AS numemployees, SUM(e.salary) as total_salaries FROM projects pro JOIN employeesprojects ep ON ep.projectid = pro.id JOIN employees e ON e.id = ep.employee_id GROUP BY project_id; `"See full answer

Aikya S. - "SELECT name ,type ,ROUND( CASE WHEN type = 'Electronic' THEN price * 0.90 WHEN type = 'Clothing' THEN price * 0.80 WHEN type = 'Grocery' THEN price * 0.95 WHEN type = 'Book' THEN price * 0.85 ELSE price END, 2 ) as discounted_price FROM products `"See full answer

Salome L. - "SELECT items.item_category, SUM(orders.orderquantity) AS totalunitsorderedlast7days FROM orders JOIN items ON orders.itemid = items.itemid WHERE orders.order_date BETWEEN DATE('now', '-6 days') AND DATE('now') GROUP BY items.item_category `"See full answer

G B. - " select user_id, b.marketing_channel from user_sessions a Left join attribution b on b.sessionid = a.sessionid group by 1,2 HAVING sum(purchasevalue)>100 and min(adclick_timestamp) `"See full answer

Anonymous Possum - "The unique id is not clear in this question"See full answer

Bhavna S. - " with youngsuccrate as( select strftime('%m', postdate) AS postmonth, round(sum(issuccessfulpost)*1.0/count(issuccessfulpost),2)as yascrate from post where userid in (select userid from post_user where age between 0 and 18) group by post_month ), nonyoungsucc_rate as( select strftime('%m', postdate) AS postmonth, round(sum(issuccessfulpost)*1.0/count(issuccessfulpost),2)as nonyasc_rate from post where user_id in (select"See full answer

Bradley E. - "I'm pretty sure Exponent's answer is wrong. In the snippet below, they use "pl.name = 'Telephones' to attempt to filter down to the Telephone transactions, but they do this within a LEFT JOIN which means all product_lines rows are returned. > LEFT JOIN product_lines pl > ON p.productlineid = pl.id > AND pl.name = 'Telephones' Below is my solution. Also, I didn't see anywhere that said the "amount" column was in cents instead of dollars, but I still divided by 100 to be consistent with Exp"See full answer

Alina G. - "Order the result in descending month is not applied in the solution"See full answer

Jaime A. - "I might be missing something but the solution, seems to be incorrect. ... , post_pairings AS ( SELECT ps.user_id, ps.postseqid AS failpostid, ps.postseqid + 1 AS nextpostid FROM post_seq AS ps WHERE ps.issuccessfulpost IS TRUE ) -- here ps.issuccessfulpost IS TRUE the condition should be FALSE -- in that way ps.postseqid is the actual failed post(failpostid) -- Additionally, at the end the join is assumming that the sequence id is going to match the post_id, wh"See full answer

Meenakshi D. - "BETWEEN and HAVING clauses in SQL serve different purposes: 1. BETWEEN Clause Used to filter rows based on a range of values. Works with numeric, date, or text values. Can be used with WHERE or HAVING clauses. The range includes both lower and upper bounds. Example: Filtering employees with salaries between 30,000 and 50,000 `SELECT * FROM Employees WHERE salary BETWEEN 30000 AND 50000;` 2. HAVING Clause Used to filter **groups"See full answer

Marcos G. - "with my_table as (select * , rownumber() over(order by customerid) as row_index from customers) select customer_id, customer_name from my_table where row_index % 3 = 0"See full answer

Derrick M. - "SELECT o.order_amount FROM orders o JOIN departments d ON d.departmentid = o.departmentid WHERE d.department_name = 'Fashion' ORDER BY order_amount DESC LIMIT 1 OFFSET 1; `"See full answer

Richard B. - "Select interface, Count(case when issuccessfulpost then 1 end) as post_success, Count() as postattempt, ROUND((COUNT(CASE WHEN issuccessfulpost THEN 1 END) * 100 / COUNT()), 2) AS postsuccess_rate from post where interface like 'Iphone%' group by 1 order by postsuccessrate desc `"See full answer

Maliki U. - "Here is my implementation: select marketing_channel, AVG(purchasevalue) as avgpurchase_value from attribution group by marketing_channel order by avgpurchasevalue DESC ; There is no need to copy and past the line of code for calculating the average into order by, just Alias is enough because going by the order of execution in sql, Always, order by is executed after executing select clause."See full answer

Victor N. - "with top_players as( Select team_id, player_name, max(gamescore) as maxscore ,denserank() over(partition by teamid order by max(game_score) desc) rk from players p join scores s on p.playerid=s.playerid group by player_name, team_id ) Select team_id, player_name, max_score from top_players where rk<=2 order by teamid, maxscore desc `"See full answer