Question

我有三张桌子

USER
---------------------------------------
id | uid | first_name | last_name | ...
---------------------------------------
0  | 95  | ...        | ...       | ...   
1  | 100 | ...        | ...       | ...     
2  | 120 | ...        | ...       | ...
3  | 130 | ...        | ...       | ...

NEWS_mm
------------------------
uid_local | uid_foreign 
------------------------
40        | 90        
40        | 100 
50        | 120 
50        | 130

NEWS
-------------------------------
id | uid  | title | image | ...
-------------------------------
0  | 40   | ...   | ...   | ...    
1  | 50   | ...   | ...   | ...
2  | 60   | ...   | ...   | ...

现在我只想从表“USER”中选择所有用户，其中新闻ID为表“NEWS”。 NEWS_mm包含来自“NEWS”的新闻ID（id_local）和来自“USER”的用户ID（id_foreign）

SELECT USER.*, NEWS_MM.*, NEWS.*
FROM USER
    JOIN NEWS_MM
        ON NEWS_MM.uid_foreign = USER.uid
    JOIN NEWS
        ON NEWS_MM.uid_local = NEWS_MM.uid_local
WHERE NEWS.uid = 50

Answer 1

你是在正确的路线，但你正在加入NEWS_MM = NEWS_MM上的新闻我怀疑你想加入这样的话：

SELECT USER.*, NEWS_MM.*, NEWS.*
FROM USER
JOIN NEWS_MM
    ON NEWS_MM.uid_foreign = USER.uid
JOIN NEWS
    ON NEWS_MM.uid_local = NEWS.uid
WHERE NEWS.uid = 50

另请注意，您知道NEWS_MM.uid_local的值与NEWS.uid相同，因此如果您只想要USER表中的数据，那么您实际上根本不需要第二次连接，这样可以提供更好的性能< / p>

SELECT 
  NEWS_MM.uid_local, 
  USER.* 
FROM USER
  JOIN NEWS_MM
    ON NEWS_MM.uid_foreign = USER.uid
WHERE NEWS_MM.uid_local = 50

MySQL - 使用关系表加入

1 个答案: