Question

我正在努力解决this HackerRank SQL coding challenge。我们有两个表，一个名为Hackers，列hacker_id和name，另一个名为Submissions，列submission_date，submission_id，{{1 }和hacker_id。

我提交的一个解决问题的查询是：

score

然而，我收回了错误：

SELECT  es.date, es.count, m.hacker_id, m.name, m.score
FROM 
(
    SELECT submission_date as date, COUNT(hacker_id) as count
    FROM (
        SELECT submission_date, COUNT(submission_id) as count, hacker_id
        FROM Submissions 
        GROUP BY submission_date, hacker_id
    ) f 
    HAVING count >=1
) es
JOIN (
    SELECT s.submission_date as date, s.hacker_id, h.name, s.score
    FROM Submissions s 
    JOIN Hackers h ON h.hacker_id = s.hacker_id 
    JOIN (SELECT submission_date, MAX(score) as score FROM Submissions GROUP BY submission_date) foo ON foo.submission_date = s.submission_date
    WHERE s.score = foo.score 
) m 
ON es.date = m.date
ORDER BY es.date

这让我很困惑，因为我在查询中没有使用聚合函数。为什么解释器会出现这个错误，我应该考虑解决它？

Answer 1

第一个子查询中有两列名为count。一个是字段submission_id的计数，第二个是hacker_id的计数。据我所知，查询需要执行submission_dates的选择，其中count大于或等于1.因此，取决于假定哪个计数大于或等于1，查询可能会被重写如下：

1）如果需要count（submission_id）大于或等于1：

SELECT  es.date, es.count, m.hacker_id, m.name, m.score
FROM 
(
SELECT submission_date as date, COUNT(hacker_id) as count
FROM (
    SELECT submission_date, COUNT(submission_id) as count, hacker_id
    FROM Submissions 
    GROUP BY submission_date, hacker_id
) f 
WHERE f.count >=1
GROUP BY submission_date
) es
JOIN (
   SELECT s.submission_date as date, s.hacker_id, h.name, s.score
   FROM Submissions s 
JOIN Hackers h ON h.hacker_id = s.hacker_id 
JOIN (SELECT submission_date, MAX(score) as score FROM Submissions GROUP BY submission_date) foo ON foo.submission_date = s.submission_date
WHERE s.score = foo.score 
) m 
ON es.date = m.date
ORDER BY es.date

2）count（hacker_id）大于或等于1：

SELECT  es.date, es.count, m.hacker_id, m.name, m.score
FROM 
(
SELECT submission_date as date, COUNT(hacker_id) as count
FROM (
    SELECT submission_date, COUNT(submission_id) as count, hacker_id
    FROM Submissions 
    GROUP BY submission_date, hacker_id
) f 
GROUP BY submission_date
HAVING count(hacker_id) >= 1
) es
JOIN (
   SELECT s.submission_date as date, s.hacker_id, h.name, s.score
   FROM Submissions s 
JOIN Hackers h ON h.hacker_id = s.hacker_id 
JOIN (SELECT submission_date, MAX(score) as score FROM Submissions GROUP BY submission_date) foo ON foo.submission_date = s.submission_date
WHERE s.score = foo.score 
) m 
ON es.date = m.date
ORDER BY es.date

问题在于子查询中使用了聚合函数计数（hacker_id）而没有＆＃34; group by＆＃34;列＃34; submission_date＆＃34;在那里定义为非聚合列。我还不确定sql-query在逻辑上是否正确。

MySQL解释器调用查询而不使用聚合函数作为聚合查询

1 个答案: