Question

我有这样的声明：

SELECT board.*,numlikes 
FROM board
LEFT JOIN (
    SELECT pins.board_id, COUNT(source_user_id) AS numlikes
    FROM likes 
    INNER JOIN pins ON pins.id = likes.pin_id 
    GROUP BY pins.board_id
) likes ON board.id = likes.board_id
WHERE who_can_tag='' 
ORDER BY numlikes DESC 
LIMIT 10

然后，我可以使用board从".$info['board_name']."中提取行。但是，我不熟悉多个连接，除了board和likes表之外，我还需要加入另一个表。第三个表格为user，并与board.user_id匹配user.user_id。

如何使用此数据从user中提取用户名？

".$info['username']."不会username在表board中搜索字段{{1}}吗？

Answer 1

这将做你想要的：

SELECT board.*,
    `user`.username,
    numlikes 
FROM board
INNER JOIN `user` ON board.user_id = `user`.user_id
LEFT JOIN (
    SELECT pins.board_id, COUNT(source_user_id) AS numlikes
    FROM likes 
    INNER JOIN pins ON pins.id = likes.pin_id 
    GROUP BY pins.board_id
) likes ON board.id = likes.board_id
WHERE who_can_tag='' 
ORDER BY numlikes DESC 
LIMIT 10

SQL多重连接整理

1 个答案: