Script to update mysql not working - php

Question

Okay so I have a PHP script that makes a user an artist if vote is high enough. The first part of the script works (the part that does the voting). However, the second part of the script that makes a user an artist does not. It worked before on localhost but is not working on live server for some reason. Either the script has changed and I didn't notice it or there's something wrong with my server config.
I know I should be using mysqli but please don't mention that I am working on it.
To explain how the system works, a form on the voting page is posted to this script and it all runs from there.
There is no error in the error log. Updating the table for //make an artist if vote high enough just doesn't work.
Here's the script:
<?php
session_start();
include("../database.php");
$username = $_SESSION["username"];
$artistname = htmlspecialchars(mysql_real_escape_string($_POST['artistname']));
$trackname = htmlspecialchars(mysql_real_escape_string($_POST['trackname']));
$trackurl = htmlspecialchars(mysql_real_escape_string($_POST['trackurl']));
$flag = 0; // Safety net, if this gets to 1 at any point in the process, we don't upload.
if(isset($_POST['yes'])){
//code runs if vote is yes
//check if user hasnt already voted on track
$result = mysql_query("SELECT username FROM voted WHERE voted='$artistname' AND trackname='$trackname' AND username='$username'")or die(mysql_error());
$check2 = mysql_num_rows($result);
if ($check2 != 0) {
echo('<t1>Sorry, you have already voted on this track. <b>Click next track.</b> </t1>');
$flag = $flag + 1;
}
//code runs if everything is okay
if($flag == 0){
mysql_query("UPDATE members SET vote = vote+1 WHERE artistname='$artistname'
");
echo '<t1><b>You liked the track "'.$trackname.'" by "'.$artistname.'"</t1></b>';
mysql_query("INSERT INTO voted (username, voted,trackname, yesno)
VALUES ('".$username."','".$artistname."','".$trackname."', 'yes')")
or die(mysql_error());
//make an artist if vote high enough
$vote = mysql_query("SELECT vote FROM members WHERE artistname='$artistname'")or die(mysql_error());
if ($vote > 50) {
$artisturl = htmlspecialchars(mysql_real_escape_string(str_replace(' ', '',$_POST['artistname'])));
mysql_query("UPDATE members SET artist='Y', image1='../files/noprofile.jpg', artisturl='$artisturl' WHERE artistname='$artistname'
")or die(mysql_error());
mysql_query("UPDATE tracks SET artist='Y', artisturl='$artisturl' WHERE artistname='$artistname'
")or die(mysql_error());
//email user that has just been made artist
$result = mysql_query("SELECT * FROM members WHERE artistname= '$artistname'");
while($row = mysql_fetch_array($result)){
function spamcheck($field)
{
//filter_var() sanitizes the e-mail
//address using FILTER_SANITIZE_EMAIL
$field=filter_var($row['email'], FILTER_SANITIZE_EMAIL);
//filter_var() validates the e-mail
//address using FILTER_VALIDATE_EMAIL
if(filter_var($row['email'], FILTER_VALIDATE_EMAIL))
{
return TRUE;
}
else
{
return FALSE;
}
}
{//send email
$to = $row['email'];
$subject = "Congratulations! You're now an NBS artist";
$message = "Hi ".$row['artistname'].",
//message removed for condensed code
$from = "";
$headers = 'From:' . "\r\n" .
'Reply-To: ' . "\r\n";
mail($to,$subject,$message,$headers);
}
}
echo '<br><t1>You just made "'.$artistname.'" an artist! <b>Click here</b> to see their profile.</t1>';
}
}
}

Answer 1

You are missing two lines to convert the resource returned by mysql_query() into an integer for the comparison with 50.
$vote = mysql_query("SELECT vote FROM members WHERE artistname='$artistname'")or die(mysql_error());
// Add these two lines
$vote = mysql_fetch_assoc($vote);
$vote = $vote['vote'];
if ($vote > 50) {
...however, all that section could be re-written to use 2 queries instead of 4:
//make an artist if vote high enough
$artisturl = mysql_real_escape_string(htmlspecialchars(str_replace(' ', '',$_POST['artistname'])));
// This effectively combines the first SELECT and the two UPDATEs into one query
$result = mysql_query("
UPDATE members m
LEFT JOIN tracks t ON m.artistname = t.artistname
SET
m.artist = 'Y',
t.artist = 'Y',
m.image1 = '../files/noprofile.jpg',
m.artisturl = '$artisturl',
t.artisturl = '$artisturl'
WHERE m.artistname = '$artistname' AND m.vote > 50
") or die(mysql_error());
// If this affected more than 0 rows, the user was made an artist
if (mysql_affected_rows($result) > 0) {
//email user that has just been made artist
$result = mysql_query("SELECT * FROM members WHERE artistname= '$artistname'");
// ...and so on
Note also that you should pass data through mysql_real_escape_string() as the last operation. So it should go mysql_real_escape_string(htmlspecialchars($data)) rather than the other way around.

Answer 2

I'll throw a dart at this one.
$vote = mysql_query("SELECT vote FROM members WHERE artistname='$artistname'")or die(mysql_error());
if ($vote > 50) {
I don't believe you are converting your mysql_query result into a useful variable. Maybe you were using mysql_fetch_assoc or mysql_num_rows ? Num rows makes more sense if you have an individual record for each vote. If you are summing them up then you can use something like
$output = mysql_fetch_assoc(mysql_query("SELECT vote FROM members WHERE artistname='$artistname'")or die(mysql_error());
$vote = $output['vote'];
Something else to point out is that you aren't using mysql_real_escape_string on your inputs. This is very dangerous and it is strongly encouraged to use this function if you are facing the public internet.