mysql从两个不同的表中选择。

Question

$sql = "select body, stamp from posts where user_id = '$userid' order by stamp desc"; 

注意：上述查询工作正常。我想要做的也是从我的用户表中选择用户名，并显示与user_id匹配的用户名。

我编辑了上述语句，所以它不起作用。有人可以建议正确的查询吗？我的目标是也能够显示用户名。除了简单地显示user_id。

$sql = "select body, stamp from posts AND username from users where user_id = '$userid' order by stamp desc"; 

我的目标是也能够显示用户名。而不是简单的user_id。

Answer 1

你需要使用JOIN带来的两个表一起在匹配字段，所以像：

$sql = "SELECT p.body, p.stamp, u.username FROM posts p INNER JOIN users u ON p.user_id=u.user_id WHERE p.user_id='$userid' ORDER BY p.stamp DESC"; 

Answer 2

$sql = "select body, stamp from posts , username from users where user_id = '$userid' order by stamp desc"; 

我只是在查询纠正语法错误...这是与关键字应该由,更换。

Answer 3

$sql = " 
select body, stamp from posts where user_id = '$userid' order by stamp desc 
UNION ALL 
select body, stamp from username where user_id = '$userid' order by stamp desc 
"; 

http://dev.mysql.com/doc/refman/5.0/en/union.html

Answer 4

select body, stamp, username 
    from posts,users 
    where users.user_id = post.user_id 
     and users.user_id = '$userid' 
    order by stamp desc; 

Answer 5

您应该使用别名或表名避免这样的重复问题

tablename.column 

或

alias.column 

您可以设置别名您的where子句中：

FROM table as alias_name 

Answer 6

$sql = "SELECT p.body,p.stamp,u.username from posts p LEFT JOIN users as u ON (p.user_id = u.user_id) WHERE user_id = '$user_id' ORDER BY p.stamp DESC"; 

应该这样做

Answer 7

您可以使用表名来选择列。例如：

$Query = "select table1.body, table1.stamp, users.username from posts, users where user_id = '$userid' order by stamp desc"; 

但是，这种方法在性能上并不好。

最好的方法是：

$Query = "SELECT table1.body, table1.stamp, users.username 
FROM posts 
INNER/LEFT/RIGHT JOIN users 
ON users.user_id = '$userid' AND users.user_stamp = stamp.stamp_id 

所有表必须进行相关。

问候，

Answer 8

试试这个：

select body, stamp, username 
from posts p JOIN users a ON p.user_id = a.user_id 
WHERE p.user_id = '$userid' order by stamp desc 

Answer 9

这应该很好地工作：

SELECT posts.body, posts.stamp, users.username 
FROM posts, users 
WHERE posts.user_id = '$userid' AND posts.user_id = users.user_id 
ORDER BY posts.stamp DESC