PHP MySQL 查询帮助

Question

我尝试使用此查询返回变量 $d['userID'] 等于单独表中的用户 ID 的每个实例，然后回显与该用户 ID 关联的用户名。

这是我到目前为止所拥有的:

$uid = $d['userID'];

$result = mysql_query("SELECT u.username
FROM users u
LEFT JOIN comments c
ON c.userID = u.id
WHERE u.id = $uid;")$row = mysql_fetch_assoc($result);
echo $row['username'];

Answer 1

这应该有效

$uid = mysql_real_escape_string($d['userID']);

$result = mysql_query("SELECT u.username
FROM users u
LEFT JOIN
 comments c
ON c.userID = u.id
WHERE u.id = '$uid'");


    while($row = mysql_fetch_assoc($result))
{
   //PRINTS ALL INSTANCES OF THE ROW
    echo $row['username'];
}

从上面的回显中打印出什么。