Question

我有3张这样的表

表1：UserInfo

#include "ioswrapper.h"
#import <UIKit/UIPasteboard.h>

ioswrapper::ioswrapper(QObject *parent):QObject(parent)
{

}
void ioswrapper::setPasteBoardString(const QString &paste)
{
        UIPasteboard *pb = [UIPasteboard generalPasteboard];
        if (pb) {
                    const char *str = paste.toUtf8().data();
                    NSString *text = [NSString stringWithCString:(const char *)str
                        encoding:(NSStringEncoding)NSUTF8StringEncoding];
                    if (text)
                        pb.string = text;
        }
}

user_id   userName

表2：帖子

123       userOne

user_id   postContent

表3：LikePost

123       This is test message

user_id   likesPostId

我想运行一个查询来获取这3个表中的帖子喜欢，帖子和用户信息的总数。

我可以为每个人执行此操作，例如在Post表中：

123       This is test message

和SELECT COUNT(*) FROM Post WHERE Post.user_id = '123'

有人在1个查询中有更好的解决方案吗？非常感谢你！

Answer 1

使用类似这样的结构化查询（带子查询）。

 SELECT u.user_id, u.userName, p.num postcount, l.num likecount
   FROM UserInfo u
   LEFT JOIN (
                 SELECT COUNT(*) num, 
                        user_id
                   FROM Post
                  GROUP BY user_id
        ) p ON u.user_id = p.user_id
   LEFT JOIN (
                 SELECT COUNT(*) num, 
                        user_id
                   FROM LikePost
                  GROUP BY user_id
        ) l ON u.user_id = l.user_id

这里发生了什么？这两个子查询，例如

                 SELECT COUNT(*) num, 
                        user_id
                   FROM LikePost
                  GROUP BY user_id

每个user_id生成一个零行或一行的虚拟表，显示每个user_id的计数。然后，将这些虚拟表连接到UserInfo表。

使用LEFT JOIN，因为普通的内置JOIN会抑制缺少帖子或喜欢的用户。

Answer 2

试试这个

SELECT ui.userName,Count(p.*),
Count(lp.*) as TotalPostLikes
 FROM UserInfo ui 
INNER JOIN Post p on p.user_id=ui.user_id
INNER JOIN LikePost lp on lp.user_id=ui.user_id
WHERE ui.user_id = '123'
GROUP BY ui.userName

如果您想在帖子上选择用户名，发布和喜欢，请尝试以下

 SELECT ui.userName,p.postContent as PostContent,
        (SELECT COUNT(lp.user_id) FROM LikePost lp 
    WHERE lp.user_id=ui.user_id) as Likes,

(SELECT COUNT(_p .user_id) FROM Post _p 
    WHERE _p .user_id=ui.user_id) as TotalPosts

     FROM UserInfo ui 
    INNER JOIN Post p on p.user_id=ui.user_id
    WHERE ui.user_id = '123'

Answer 3

是的，您可以使用leftjoin上的Post和下面的LikePost在一个查询中执行此操作

SELECT COUNT(*),User.userName FROM UserInfo as User
leftjoin  Post as Post on Post.user_id = User.user_id
leftjoin  LikePost as LikePost on LikePost.user_id = User.user_id      
where Post.user_id = 123
group by Post.user_id

如何从3个具有相同用户ID的表中获取数据？

3 个答案: