警告: mysql_result() [function.mysql- result ]：无法跳到中MySQL结果索引6的第0行

Question

我有一些代码，

if(logged_in() == true){ 
$website_id = (INT)$_GET['id'];
$addedById  = mysql_result(mysql_query("SELECT `user_id` FROM `websites` WHERE `id` = '$website_id'"), 0);
$q = mysql_fetch_assoc(mysql_query("SELECT `phone` FROM `users` WHERE `user_id` = '$addedById'"));
$result       = mysql_query("SELECT * FROM `websites` WHERE `id` = '$website_id'");
echo $q['phone'];
 }
else{ 
echo' please logged_in'; 
 }

并返回错误“Warning: mysql_result() function.mysql-result: Unable to jump to row 0 on MySQL result”

在行中的

$addedById  = mysql_result(mysql_query("SELECT `user_id` FROM `websites` WHERE `id` = '$website_id'"), 0);

我尝试了几种方法来解决这个问题，但都没有结果，我该如何解决它呢？

Answer 1

尝试分离代码并检查查询结果。

$result = mysql_query('SELECT user_id FROM websites WHERE id = '.$website_id);
if (!$result || empty($result)) {
    echo 'Could not query:' . mysql_error();
}
echo mysql_result($result, 0);

以下是经过一些更改的代码(不是一个很好的代码，但可以尝试一下)：

if(logged_in() == true){ 
    $website_id = (INT)$_GET['id'];

    $result = mysql_query("SELECT user_id FROM websites WHERE id = " . $website_id);
    if (!empty($result)) {
        $addedById = mysql_result($result, 0);

        $query_user = mysql_query("SELECT phone FROM users WHERE user_id = " . $addedById);
        $result_user = mysql_fetch_assoc($query_user);
        if (!empty($result_user) {
            echo $result_user[0]['phone'];
        }
    }
 } else { 
    echo' please logged_in'; 
 }

但是您不应该再使用此mysql函数，它们已被弃用。使用PDO或MySQLi。http://php.net/manual/en/book.mysqli.php http://php.net/manual/en/ref.pdo-mysql.php