我想使用 MySQl 将多个表结果计数合并到一个结果中答案

【问题标题】：I want to combine multiple tables result count in one result using MySQl我想使用 MySQl 将多个表结果计数合并到一个结果中
【发布时间】：2012-04-27 07:43:52
【问题描述】：

我想使用 MySql（查看所需结果）将多个表的结果计数合并到一个结果中，但如果您有任何疑问，我会对此感到困惑或对此进行优化的方法，请回答我或帮助绝对感谢我自己也尝试过这个 Query 与 Union 但严重没有成功（请参阅查询示例 em>)

需要结果

post  post_comment_likes  post_comments  post_likes
2     0                   3              0

查询

SELECT COUNT(*) AS `post` FROM post WHERE user_id = "123456" UNION
SELECT COUNT(*) AS `post_comment_likes` FROM post_comment_likes WHERE user_id = "123456" UNION 
SELECT COUNT(*) AS `post_comments` FROM post_comments WHERE user_id = "123456" UNION
SELECT COUNT(*) AS `post_likes` FROM post_likes WHERE user_id = "123456"

【问题讨论】：

标签： mysql database optimization logic

【解决方案1】：

set @post = 0;
set @post_comment_likes = 0;
set @post_comments = 0;
Set @post_likes = 0;


 select count(*) into @post from .......;
 select count(*) into @post_comment_likes from ......;
 select count(*) into @post_comments from ......;
 select count(*) into @post_likes from ......;

select (@post + @post_comment_likes + @post_comments + @post_likes) ;

在上面的代码中，@post、@post_comment_likes、@post_cmets、@post_likes 是会话变量，你需要在它们前面加上'@'字符
使用上面的代码，您可以访问每个表的计数。

【讨论】：

【解决方案2】：

SELECT (SELECT COUNT(*) 
          FROM post 
         WHERE user_id = "123456") AS `post`
      ,(SELECT COUNT(*) 
          FROM post_comment_likes
         WHERE user_id = "123456") AS `post_comment_likes`
      ,(SELECT COUNT(*)
          FROM post_comments
         WHERE user_id = "123456") AS `post_comments`
      ,(SELECT COUNT(*)
          FROM post_likes
         WHERE user_id = "123456") AS `post_likes`

您应该将COUNT(*) 替换为COUNT(<some fieldname>)，并且可以为用户ID 使用变量。

【讨论】：

【解决方案3】：

请尝试下面给出的mysql查询

SELECT
(SELECT COUNT(*) AS `post` FROM post WHERE user_id = "123456" ) AS post , 
(SELECT COUNT(*) AS `post_comment_likes` FROM post_comment_likes WHERE user_id ="123456") AS post_comment_likes,
(SELECT COUNT(*) AS `post_comments` FROM post_comments WHERE user_id = "123456") AS post_comments , 
(SELECT COUNT(*) AS `post_likes` FROM post_likes WHERE user_id = "123456" ) AS post_likes

谢谢

【讨论】：