Question

这是一个复杂的查询，我希望能够实现一个语句，而不是在PHP中处理数组值。

达到所需的输出：

User    Jobs    Total
John D. 5       $1245.67
Mary L. 3       $800.56

到目前为止，这是我的查询：

SELECT
  SUM(job.cost) AS sum,
  COUNT(DISTINCT job.user) as count,
  user.id, user.firstname, user.lastname
FROM      `job`
LEFT JOIN `user` ON job.user = user.id
GROUP BY user.id

但计数值是错误的：它是不同的用户，所以当然每个都是错的。我该如何解决这个问题？

表user

id, name, etc.

表job

id, user, cost

一个user到多个job

更新

这似乎正常：

SELECT 
  SUM(job.cost) AS sum,
  COUNT(1) as count,
  user.id, user.firstname, user.lastname
FROM `job`
LEFT JOIN `user` ON job.user = user.id
GROUP BY user.id

Answer 1

SET @seq = 0;

SELECT place FROM

(SELECT
  @seq := @seq + 1 AS place
  SUM(job.cost) AS sum,
  COUNT(1) as count,
  user.id, user.firstname, user.lastname
FROM      `job`
LEFT JOIN `user` ON job.user = user.id
GROUP BY user.id
ORDER BY COUNT(1) DESC) 

AS list
WHERE list.id = 'my_user_id'

MySQL Join plus Count plus Sum

1 个答案: