下面是创建我的表的脚本:

CREATE TABLE clients (
   client_i INT(11),
   PRIMARY KEY (client_id)
);
CREATE TABLE projects (
   project_id INT(11) UNSIGNED,
   client_id INT(11) UNSIGNED,
   PRIMARY KEY (project_id)
);
CREATE TABLE posts (
   post_id INT(11) UNSIGNED,
   project_id INT(11) UNSIGNED,
   PRIMARY KEY (post_id)
);

在我的PHP代码中,当删除客户端时,我想删除所有项目的帖子:

DELETE 
FROM posts
INNER JOIN projects ON projects.project_id = posts.project_id
WHERE projects.client_id = :client_id;

posts表没有外键client_id,只有project_id。我想删除具有传递client_id的项目中的帖子。

这是不工作的,因为没有帖子被删除。


当前回答

试试下面的方法:

DELETE posts.*,projects.* 
FROM posts
INNER JOIN projects ON projects.project_id = posts.project_id
WHERE projects.client_id = :client_id;

其他回答

试试下面的方法:

DELETE posts.*,projects.* 
FROM posts
INNER JOIN projects ON projects.project_id = posts.project_id
WHERE projects.client_id = :client_id;

使用子选择的另一种删除方法是WHERE EXISTS,这种方法比使用IN更好

DELETE  FROM posts
WHERE   EXISTS ( SELECT  1 
                 FROM    projects
                 WHERE   projects.client_id = posts.client_id);

使用这个而不是连接的一个原因是带有join的DELETE禁止使用LIMIT。如果希望以块的形式进行删除,从而不产生完整的表锁,则可以使用delete WHERE EXISTS方法添加LIMIT。

试试这个,

DELETE posts.*
FROM posts
INNER JOIN projects ON projects.project_id = posts.project_id
WHERE projects.client_id = :client_id

如果join不适合你,你可以尝试这个解决方案。它用于在不使用外键+特定的where条件时从t1中删除孤立记录。也就是说,它从table1中删除有空字段“code”而在table2中没有记录的记录,通过字段“name”匹配。

delete table1 from table1 t1 
    where  t1.code = '' 
    and 0=(select count(t2.name) from table2 t2 where t2.name=t1.name);

你也可以像这样使用别名,它只是在我的数据库上使用它!T是需要删除的表!

DELETE t FROM posts t
INNER JOIN projects p ON t.project_id = p.project_id
AND t.client_id = p.client_id