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($login_query, 0, 'user_id') : false;
}
最佳答案
您的 SQL 语句中需要一个 GROUP BY 子句。
"SELECT COUNT(`user_id`) as `count`,`user_id`
FROM `users`
WHERE `email`='$email'
AND `password`='".md5($password)."'
GROUP BY `user_id`"
关于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,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10253195/