mysql從兩個不同的表中選擇。

Question

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

注意：上面的查詢工作正常。 我還想從用戶表中選擇用戶名，並顯示與user_id匹配的用戶名。

我已經像上面那樣編輯了上面的語句，但它不起作用。 有人可以提出正確的查詢嗎？ 我的目標是也能夠顯示用戶名。 除了簡單地顯示user_id。

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

我的目標是也能夠顯示用戶名。 而不是簡單的user_id。

Answer 1

您需要使用JOIN在匹配字段上將兩個表放在一起，如下所示：

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

Answer 2

您可以使用表名稱選擇列。 例如：

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

但是，這種方法的性能不好。

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

所有表必須相關。

問候，

Answer 3

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

我只是在查詢糾正語法錯誤......那是應改為AND關鍵字, 。

Answer 4

$sql = "
select body, stamp from posts where user_id = '$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 5

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

Answer 6

您應該使用別名或表名來避免類似的重復問題

tablename.column

要么

alias.column

您可以在where子句中設置別名：

FROM table as alias_name

Answer 7

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

應該做

Answer 8

嘗試這個：

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

這應該工作正常：

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