Question

我正在撞墙，试图弄清楚这一个。我有一个2路左连接查询工作完美：

'SELECT f.user_id, u.avatar
    FROM following AS f
    LEFT JOIN users AS u ON u.username = f.user_id
    WHERE f.follower_id = "'.$u.'"';

从一个像桌子这样的推特，然后从一个单独的成员表中添加他们的头像，从而拉出一个成员列表。

现在我正在尝试从第三个表（实际的推文）中提取所有数据，但它不起作用：

'SELECT f.user_id, u.avatar, p.*
    FROM following AS f
    LEFT JOIN users AS u ON u.username = f.user_id
    WHERE f.follower_id = "'.$u.'"      
    LEFT JOIN posts AS p ON p.username = u.username';

我也尝试过添加：

WHERE p.username = "'.$u.'"';

到最后，但它并没有什么区别。该错误只是说第二个左连接存在SQL语法错误。关于我缺少什么的任何建议？

Answer 1

join是from子句的一部分。 where子句跟在from：

之后

SELECT f.user_id, u.avatar, p.*
FROM following AS f
LEFT JOIN users AS u ON u.username = f.user_id
LEFT JOIN posts AS p ON p.username = u.username
WHERE f.follower_id = "'.$u.'"    ;

Answer 2

你需要在加入之后使用where条件：

'SELECT f.user_id, u.avatar, p.*
    FROM following AS f
    LEFT JOIN users AS u ON u.username = f.user_id

    LEFT JOIN posts AS p ON p.username = u.username  WHERE f.follower_id = "'.$u.'"     ' ;

Answer 3

将WHERE子句移动到最后一次LEFT JOIN之后。

SELECT f.user_id, u.avatar, p.*
FROM following AS f
LEFT JOIN users AS u ON u.username = f.user_id    
LEFT JOIN posts AS p ON p.username = u.username
WHERE f.follower_id = "'.$u.'";

另外，出于安全原因，您可能希望使用预准备语句，而不是将$ u直接连接到查询。

3路左连接查询错误

3 个答案: