Mysql_result() , Error

[TottoBennington]

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/tottoswd/public_html/iuw/func/user.func.php on line 4

 

<?php
function user_exists($email) {
$email = mysql_real_escape_string($email);
$query = mysql_query("SELECT COUNT(´user_id´) FROM ´users_vu´ WHERE ´email´ = '$email'");
return (mysql_result($query, 0) == 1) ? true : false;    (THIS IS THE LINE WITH THE ERROR)
?>

[dragon_sa]

I believe this is an issue with the statement try changing it to this

$query = mysql_query("SELECT COUNT(user_id) FROM users_vu WHERE email = '$email'");

[cpd]

Whether you use the slanted apostrophe or not shouldn't matter. The correct syntax and good practice is to use the apostrophe;  but as I said, is not required.

 

I believe the error is due to your "COUNT" function. Try putting am as statement and accessing the alias: "COUNT(*) AS Total".

 

$query = mysql_query("SELECT COUNT(´user_id´) AS `User` FROM ´users_vu´ WHERE ´email´ = '$email'");
return (mysql_result($query, 0, 'User') == 1) ? true : false;    (THIS IS THE LINE WITH THE ERROR)

 

[scootstah]

The problem is that you are using some other kind of apostrophe and not a back-tick. This is correct:

$query = mysql_query("SELECT COUNT(`user_id`) FROM `users_vu` WHERE `email` = '$email'");

[cpd]

Oh dear, my failure to pay attention to detail :|.