SQL Interview Questions

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

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

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

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

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

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

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

Shriganesh K. - "not sure what's wrong here> select a.marketing_channel, avg(purchasevalue) as avgpurchase_value, sum(case when a.purchasevalue > 0 then 1 else 0 end) * 1.0 /count(a.sessionid) as conversion_rate from attribution a left join usersessions u on a.sessionid = u.session_id group by a.marketing_channel order by conversion_rate desc `"See full answer

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

Ali H. - "SQL databases are relational, NoSQL databases are non-relational. SQL databases use structured query language and have a predefined schema. NoSQL databases have dynamic schemas for unstructured data. SQL databases are vertically scalable, while NoSQL databases are horizontally scalable."See full answer

MnM - "SELECT c.customerid, c.customername FROM customers c WHERE rowid%3=0 `"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

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

Lin R. - "Why can not group by only on name? it gave me incorrect results when I try to do that."See full answer

Alina G. - "SELECT upsellcampaignid, COUNT(DISTINCT trans.userid) AS eligibleusers FROM campaign JOIN "transaction" AS trans ON transactiondate BETWEEN datestart AND date_end JOIN user ON trans.userid = user.userid WHERE iseligibleforupsellcampaign = 1 GROUP BY upsellcampaignid `"See full answer