如何通过另一个选择结果使用select命令?(mysql、php)
我有一些问题 这是我的样品表如何通过另一个选择结果使用select命令?(mysql、php),php,mysql,sql,Php,Mysql,Sql,我有一些问题 这是我的样品表 create table Board ( readLevel tinyint not null, writeLevel tinyint not null, PRIMARY KEY (boardID) ) engine=InnoDB character set=utf8; create table Post ( postID int not null AUTO_INCREMENT, title char(50) not n
create table Board (
readLevel tinyint not null,
writeLevel tinyint not null,
PRIMARY KEY (boardID) ) engine=InnoDB character set=utf8;
create table Post (
postID int not null AUTO_INCREMENT,
title char(50) not null,
content TEXT not null,
writeDate date not null,
readCount int not null,
PRIMARY KEY (postID)) engine=InnoDB character set=utf8;
create table Save_Board_Post(
boardID char(30) not null,
postID int not null,
FOREIGN KEY (boardID) REFERENCES Board(boardID) ON DELETE CASCADE ON UPDATE CASCADE,
FOREIGN KEY (postID) REFERENCES Post(postID) ON DELETE CASCADE ON UPDATE CASCADE ) engine=InnoDB character set=utf8;
第一张桌子是非常简单的木板,第二张桌子是柱子
三是董事会与岗位的关系表
我想在Post表中使用SELECT命令,从“保存板”中选择结果
e、 g
在这种情况下,控制台中的输出结果是
张贴
一,
二,
我想在下一次选择中使用这个结果
select * from Post where (SELECT Result);
如何放置它们
我已经用过这个句子了
select * from Post where "select postID from Save_Board_Post where boardID= 'testBoard' ";
但在控制台中没有结果
请让我知道。谢谢 您要使用的称为子查询
select *
from Post where PostID IN
(select postID
from Save_Board_Post
where boardID= 'testBoard');
SELECT * FROM Post p
WHERE p.postID IN (
SELECT postID FROM Save_Board_Post
WHERE boardID = 'testboard');
哦,你跑得真快!谢谢你的回答!非常感谢你!我猜条件句必须用引号括起来,而不是放在括号里。再次感谢@波罗罗罗卡你可以选择我正确的答案来感谢我。
SELECT * FROM Post p
WHERE p.postID IN (
SELECT postID FROM Save_Board_Post
WHERE boardID = 'testboard');