Question

SELECT contest.contest_id 
FROM db_user
LEFT JOIN participants ON participants.user_id = db_user.db_user
LEFT JOIN contest ON participants.contest_id = contest.contest_id
WHERE contest.user_id=$user
group by participants.contest_id,participants.user_id";

这将返回所有参与竞赛并且运行良好的用户，但我需要找到它返回的行数。例如，上面返回5行我希望结果为

Contest_Id       Number of users
1                       3
2                       4

Answer 1

您需要使用COUNT聚合函数以及相应的HAVING子句。请参阅以下内容：HAVING Clause

Answer 2

试试这个：

 select contest.contest_id ,Count(*) as 'No of Users'
    FROM db_user
    LEFT JOIN participants ON participants.user_id = db_user.db_user
    LEFT JOIN contest ON participants.contest_id = contest.contest_id
    WHERE contest.user_id=$user
    group by participants.contest_id;

Answer 3

尝试：

    SELECT contest.contest_id, 
(select count(*) from db_user 
      inner JOIN participants ON participants.user_id = db_user.user_id and                   
      participants.contest_id = contest.contest_id)
    FROM contest

Answer 4

您可能希望扩展SELECT语句以包含COUNT(participants.user_id)。如果用户两次参加相同的比赛，他将被重复计算。

Answer 5

您的查询在语法上是错误的，因为您应该始终选择GROUP BY子句中提到的字段，尽管MySQL允许您指定其他字段，但您应该知道大多数情况下您将获得不可预测的结果。话虽如此，也许这是您正在寻找的查询：

SELECT participants.contest_id, COUNT(DISTINCT participants.user_id) AS 'Number of users'
FROM db_user
  LEFT JOIN participants ON participants.user_id = db_user.db_user
  LEFT JOIN contest ON participants.contest_id = contest.contest_id
WHERE contest.user_id = $user
GROUP BY participants.contest_id;

MYsql查询疑难解答

5 个答案: