Question

我已经挑选出在不同难度级别的挑战中获得满分的人。但是，该问题指出，要查询hacker_id和不止一次获得满分的人员的姓名。我遇到COUNT个问题。我试图计算出现在表格中的名字的出现频率，但是我不允许这样做。我怀疑GROUPBY语法有问题。有人可以帮我吗？

以前的代码

Select s.challenge_id,s.hacker_id,h.name,s.submission_id,c.difficulty_level,s.score
FROM (((Hackers AS h JOIN Submission AS s ON h.hacker_id=s.hacker_id)JOIN Challenges AS c ON c.challenge_id=s.challenge_id)JOIN Difficulty AS d ON d.difficulty_level=c.difficulty_level)
WHERE c.difficulty_level=d.difficulty_level and s.score=d.score

结果

 challenge_id | hacker_id | name |submission_id |difficulty_level |score

    71055        86870      Todd     94613                 2    30
    66730        90411      Joe      97397                 6    100
    71055        90411      Joe      97431                 2    30

问题

 Select g.hacker_id,g.name,COUNT(g.name)
 FROM (Select s.challenge_id,s.hacker_id,h.name,s.submission_id,c.difficulty_level,s.score
 FROM (((Hackers AS h JOIN Submission AS s ON h.hacker_id=s.hacker_id)JOIN Challenges AS c ON c.challenge_id=s.challenge_id)JOIN Difficulty AS d ON d.difficulty_level=c.difficulty_level)
 WHERE c.difficulty_level=d.difficulty_level and s.score=d.score) AS g
 WHERE COUNT(g.name)>1
 GROUBY g.hacker_id,g.name;

Answer 1

如果您需要第一个查询中多次出现的每个名称的COUNT，则可以使用以下内容

SELECT hacker_id, name, COUNT(name)
FROM
(
Select s.challenge_id,s.hacker_id,h.name,s.submission_id,c.difficulty_level,s.score
FROM (((Hackers AS h JOIN Submission AS s ON h.hacker_id=s.hacker_id)JOIN Challenges AS c ON c.challenge_id=s.challenge_id)JOIN Difficulty AS d ON d.difficulty_level=c.difficulty_level)
WHERE c.difficulty_level=d.difficulty_level and s.score=d.score
) AS T
GROUP BY hacker_id, name
HAVING COUNT(name) > 1

HAVING用于过滤汇总结果

从数据库表获取值时出错

1 个答案: