"SELECT DISTINCT title,
ROUND(AVG(rating) over (partition by title),1) avg_rating,
ROUND(AVG(rating) over (partition by genre),1) genre_rating
FROM rating r
JOIN movie m
ON r.movieid=m.movieid
ORDER by 1"
Harshi B. - "SELECT DISTINCT title,
ROUND(AVG(rating) over (partition by title),1) avg_rating,
ROUND(AVG(rating) over (partition by genre),1) genre_rating
FROM rating r
JOIN movie m
ON r.movieid=m.movieid
ORDER by 1"See full answer
"WITH CTE AS (
SELECT *, ROWNUMBER()OVER(PARTITION BY utxoid ORDER BY transactionid) AS trxrk
FROM transactions
JOIN transaction_inputs
USING (transaction_id)
JOIN utxo
USING (utxo_id)
)
SELECT transaction_id AS InvalidTransactionId
FROM CTE
WHERE sender!=address OR trx_rk > 1
`"
E L. - "WITH CTE AS (
SELECT *, ROWNUMBER()OVER(PARTITION BY utxoid ORDER BY transactionid) AS trxrk
FROM transactions
JOIN transaction_inputs
USING (transaction_id)
JOIN utxo
USING (utxo_id)
)
SELECT transaction_id AS InvalidTransactionId
FROM CTE
WHERE sender!=address OR trx_rk > 1
`"See full answer
"this task is misleading . i used lag(1) and lead(1) cuz it did not say "compare temperature from 2 days before and 1 day before" , it reads to me as if its asking "compare cur temperature to prev and future and see if it rose and fall""
Erjan G. - "this task is misleading . i used lag(1) and lead(1) cuz it did not say "compare temperature from 2 days before and 1 day before" , it reads to me as if its asking "compare cur temperature to prev and future and see if it rose and fall""See full answer
"with cte as
(select *,
row_number() over(order by score desc) as rn
from players)
select player_name, score, rn as ranking
from cte
where rn= 4 or rn =6 or rn =11
`"
Gowtami K. - "with cte as
(select *,
row_number() over(order by score desc) as rn
from players)
select player_name, score, rn as ranking
from cte
where rn= 4 or rn =6 or rn =11
`"See full answer
Data Scientist
SQL
+1 more
🧠 Want an expert answer to a question? Saving questions lets us know what content to make next.
"SELECT
e1.empid AS manageremployee_id,
e1.empname AS managername,
COUNT(e2.empid) AS numberofdirectreports
FROM employees AS e1
INNER JOIN employees AS e2
ON e2.managerid = e1.empid
GROUP BY e1.emp_id
HAVING COUNT(e2.emp_id) >= 2
ORDER BY numberofdirectreports DESC, managername ASC
`"
Alvin P. - "SELECT
e1.empid AS manageremployee_id,
e1.empname AS managername,
COUNT(e2.empid) AS numberofdirectreports
FROM employees AS e1
INNER JOIN employees AS e2
ON e2.managerid = e1.empid
GROUP BY e1.emp_id
HAVING COUNT(e2.emp_id) >= 2
ORDER BY numberofdirectreports DESC, managername ASC
`"See full answer
"SELECT COUNT(DISTINCT o.customerid) AS customers, d.departmentname
FROM orders o
INNER JOIN departments d
ON d.departmentid = o.departmentid
WHERE d.departmentname IN ('Electronics','Fashion') AND o.orderdate BETWEEN '2022-01-01' AND '2022-12-31'
GROUP BY d.department_name;
`"
Derrick M. - "SELECT COUNT(DISTINCT o.customerid) AS customers, d.departmentname
FROM orders o
INNER JOIN departments d
ON d.departmentid = o.departmentid
WHERE d.departmentname IN ('Electronics','Fashion') AND o.orderdate BETWEEN '2022-01-01' AND '2022-12-31'
GROUP BY d.department_name;
`"See full answer
"SELECT
p1.player_name AS player1,
p2.player_name AS player2,
ABS(p1.level - p2.level) AS level_disparity
FROM
players p1
JOIN
players p2 ON p1.playername < p2.playername
WHERE
ABS(p1.level - p2.level) <= 5
ORDER BY
level_disparity ASC;"
Jayveer S. - "SELECT
p1.player_name AS player1,
p2.player_name AS player2,
ABS(p1.level - p2.level) AS level_disparity
FROM
players p1
JOIN
players p2 ON p1.playername < p2.playername
WHERE
ABS(p1.level - p2.level) <= 5
ORDER BY
level_disparity ASC;"See full answer