php - 第 9 行警告 : mysql_result(): supplied argument is not a valid MySQL result resource in/home/a5751969/public_html/func/user. func.php

Question

我看了很多遍代码并重写了它，但仍然无法弄清楚如何修复错误。

有第6-10行；

function login_check ($email, $password) {
$email = mysql_real_escape_string ($email);
$login_query = mysql_query("SELECT COUNT(`user_id`) as `count`,`user_id` FROM `users` WHERE `email`='$email' AND `password`='".md5($password)."'");
return (mysql_result($login_query, 0) == 1) ? mysql_result($login_query, 0, 'user_id') : false;

重写代码；

function login_check ($email, $password) {
    $email = mysql_real_escape_string($email);
    $login_query = mysql_query("SELECT COUNT(`user_id`) as `count`, `user_id` FROM `users` WHERE `email`='$email' AND `password`='".md5($password)."'");
    return (mysql_result($login_query, 0) == 1) ? mysql_result($login_query, 0, 'user_id') : false; 
}

Answer 1

您的 SQL 语句中需要一个 GROUP BY 子句。

"SELECT COUNT(`user_id`) as `count`,`user_id` 
    FROM `users` 
    WHERE `email`='$email' 
        AND `password`='".md5($password)."'
    GROUP BY `user_id`"