Question

我有三张这样的表：

漫画

|   id  | sendby
------------------
|   1   |   1
|   2   |   1
|   3   |   1
|   4   |   1

chaprer

|   id  | sendby    | translator | graphic
-------------------------------------------
|   1   |   1       |   admin    |  other
|   2   |   1       |   admin    |  other
|   3   |   1       |   admin    |  admin
|   4   |   1       |   other    |  admin

用户

|   userid  | username  
-----------------------
|      1    |   admin

我正在尝试计算我发送的内容，但结果不正确。

我试过以下查询：

SELECT username,
   SUM(
       CASE WHEN m.sendby = u.userid
       THEN 1 ELSE 0 END
   ) AS manga,
   SUM(
       CASE WHEN c.sendby = u.userid
       THEN 1 ELSE 0 END
   ) AS chapter,
   SUM(
       CASE WHEN c.translator = u.username
       THEN 1 ELSE 0 END
   ) AS translator,
   SUM(
       CASE WHEN c.graphic = u.username
       THEN 1 ELSE 0 END
   ) AS graphic
FROM user u
    left JOIN manga m 
        ON m.sendby = u.userid 
    left JOIN chapter c 
        ON c.sendby = u.userid 
where u.userid = '1'

但它返回的是不正确的：

漫画：16，第16章，译者：12，图：8

Answer 1

以下是此

的查询

select
  m.id,
  count(m.sendby) as Manga,
  lc.Chapter,
  rc.Translater,
  c.Grafix
from manga as m
  left join (select sendby, count(sendby) as Chapter from chaprer) as lc on lc.sendby  = m.sendby
  left join (select sendby, count(translater) as Translater from chaprer where translater = 'admin') as rc on rc.sendby  = m.sendby
  left join (select sendby, count(graphics) as Grafix from chaprer where translater = 'admin') as c on c.sendby  = m.sendby

编辑
我已经测试了这个查询，这就得到了这个结果

Query Result
id  Manga   Chapter Translater  Grafix
1   4       4       3           3

Answer 2

您是否故意使用left JOIN？你知道吗works怎么样？使用它两次使您的选择运行在16行结果表上。首先尝试选择所有结果（*），然后修复FROM子句以返回所需的数据以供选择。

根据您的评论，您似乎可以使用Inner join。

Answer 3

直接的方法是做这样的事情：

SELECT  username,
(SELECT COUNT(m.sendby) FROM manga m WHERE m.sendby= u.userid) AS manga,
(SELECT COUNT(c.sendby) FROM chapter c WHERE c.sendby= u.userid) AS chapter,
(SELECT COUNT(t.sendby) FROM chapter t WHERE t.translator= u.username) AS translator,
(SELECT COUNT(g.sendby) FROM chapter g WHERE g.graphic= u.username) AS graphic
FROM USER u
WHERE userid=1

如果您运行，您将在原始查询中看到问题：

SELECT *
FROM USER u
    LEFT JOIN manga m 
        ON m.sendby = u.userid 
    LEFT JOIN chapter c 
        ON c.sendby = u.userid 
WHERE u.userid = '1'

左连接返回重复的行。

一次查询多次计数

3 个答案: