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

试试这个,

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

其他回答

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。

注意,你不能在需要删除的表上使用别名

DELETE tbl_pagos_activos_usuario
FROM tbl_pagos_activos_usuario, tbl_usuarios b, tbl_facturas c
Where tbl_pagos_activos_usuario.usuario=b.cedula
and tbl_pagos_activos_usuario.cod=c.cod
and tbl_pagos_activos_usuario.rif=c.identificador
and tbl_pagos_activos_usuario.usuario=c.pay_for
and tbl_pagos_activos_usuario.nconfppto=c.nconfppto
and NOT ISNULL(tbl_pagos_activos_usuario.nconfppto)
and c.estatus=50

试试这个,

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

或者是同样的东西,只是语法略有不同(IMO更友好):

DELETE FROM posts 
USING posts, projects 
WHERE projects.project_id = posts.project_id AND projects.client_id = :client_id;

顺便说一句,在mysql中使用连接几乎总是比子查询更快…

mysql> INSERT INTO tb1 VALUES(1,1),(2,2),(3,3),(6,60),(7,70),(8,80);

mysql> INSERT INTO tb2 VALUES(1,1),(2,2),(3,3),(4,40),(5,50),(9,90);

从一个表中删除记录:

mysql> DELETE tb1 FROM tb1,tb2 WHERE tb1.id= tb2.id;

删除记录从两个表:

mysql> DELETE tb2,tb1 FROM tb2 JOIN tb1 USING(id);