php - 从一个表中选取行，但 where 语句条件在另一表中

Question

我有2个表，我必须使用where =“this value in some other table”来获取一个表的数据

users_info and users_frnds

users_info look like this

   name          image       presently      id
 somename      somimage         studying     2
 somename      somimage         studying     3

users_frnds table looks like this

  userid     friendid
     1          2
     1          3

$query = "SELECT * FROM users_info WHERE users_info.id =
users_frnds.friendid";
 
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)){
echo $row['name']. " - ". $row['image'];
echo "<br />";

但这里似乎不起作用。我想一次将所有数据放入我的数组中。

它引发了我这个错误:

Unknown column 'users_frnds.friendid' in 'where clause'

Answer 1

您需要加入，例如:

SELECT specifyfields 
FROM users_friends 
  INNER JOIN users_info ON users_info.id=users_friends.friendid

然后您将获得访问权限，尽量不要在 WHERE 子句中连接表，因为这可能会创建非常不可读的查询。