Why isn't my code picking results from the mysql?

JustCoding

Alright, I know the data is in the mysql database. I can not for the life of me figure out why my code keeps giving me a return of 0 when selecting information in this particular code. I have been selecting data from my database in another table good up until now. Do you guys see anything wrong with this code that I may have over look

Please don't preach to me about how I should be using mysqli or pdo. All good things come in time.

[ATTACH=CONFIG]4885[/ATTACH]

<?php
include 'core/functions/videoupload.php';
$youtubeid = ($_GET['id']); 

//Setting Youtube Video From Database

$videoquery = mysql_query("SELECT * FROM `video` WHERE `youtubeid` = '$youtubeid'");
if ($videoquery == 0) {
    echo 'Sorry this video does not exist in our library!';
    exit;
}else{

while ($row = mysql_fetch_row($videoquery)){
$dbvideoid = $row['0']; //Get Video ID
$dbvideorep = $row['1']; //Get Video REP
$dbyoutubeid = $row['2']; //Get Video Youtube ID

}
echo '<iframe width="560" height="315" src="http://www.youtube.com/embed/'.$dbyoutubeid.'" frameborder="0" allowfullscreen></iframe>';
}
?>

gGRwG7S.png

JustCoding

I even ran a query inside of MYSQL to see if the code worked and it works. I think the problem is $videoquery is not setting itself to a value.

JustCoding

Okay, I added or die(mysql_error()); to the end of the code and it shows the error No database selected. So, I added the code:

mysql_select_db( 'gameplayhr' );

to the top of my script, but its still not working

barand

Firstly, mysql_query() returns false if it fails, so you need to test for that value.

Secondly, finding no records is not a failure so use mysql_num_rows() to see if any were found.

laserlight

JustCoding wrote:
Please don't preach to me about how I should be using mysqli or pdo. All good things come in time.

Nice pre-emptive strike, but it doesn't work: you should be using MySQLi or PDO so that you can use prepared statements where appropriate. As it stands, your code is susceptible to SQL injection because you embed the value of $youtubeid directly in the SQL statement string before sanitising it. All good things come in time, and the time is now.

traq

JustCoding;11028029 wrote:
...its still not working

This is very unhelpful. If it were working, you probably wouldn't still be asking for help.

So, what do you mean by "not working"? Are you getting another error? Is the query returning an unexpected result? Please describe your problem in detail.

NogDog

This does not tell you how many videos it found, only whether or not MySQL was able to parse and execute the query:

$videoquery = mysql_query("SELECT * FROM `video` WHERE `youtubeid` = '$youtubeid'");
if ($videoquery == 0) {
    echo 'Sorry this video does not exist in our library!';
    exit;
}

I like to define the query string first as a variable, so that I can output it with the debug info if needed. You can then use mysql_num_rows() to determine if it found anything at all:

$sql = "SELECT * FROM `video` WHERE `youtubeid` = '".mysql_real_escape_string($youtubeid)."'";
$videoquery = mysql_query($sql);
if($videoquery == false) {
    error_log(mysql_error().PHP_EOL.$sql);
    die("Sorry, there was an unexpected DB error. Debug data has been logged");
}
elseif(mysql_num_rows($videoquery) == 0 ) {
    echo 'Sorry this video does not exist in our library!';
    exit;
}
else {