What is wrong with my script?

[noisyscanner]

My script:

^<?php

^{$game_id = $_GET['id'];}

^{mysql_connect('localhost','username','pass');}

^{mysql_select_db('database');}

^{$result = mysql_query('select * from `games` where ID = $id');}

^{while ($row = mysql_fetch_array($result, MYSQL_BOTH))}

^{

^{echo $row['game'];}

^}

^?>

Is returning error:

_{Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/brad291/public_html/games/play.php on line 6}

What is wrong with my script?

[Maq]

You need to use double quotes in the query, not single or the $id variable won't interpolate.

 

$result = mysql_query("select * from `games` where ID = $id");

[AmandaF]

In addition to what Maq said, it looks like you're calling the same variable $game_id in one spot and $id in another.  (Or are they supposed to be two separate variables?)

[Maq]

In addition to what Maq said, it looks like you're calling the same variable $game_id in one spot and $id in another.  (Or are they supposed to be two separate variables?)

 

Nice catch, yes, the OP probably wants:

 

$result = mysql_query("select * from `games` where ID = '$game_id'");

 

Note: If field ID in your table is an integer then you don't need the single quotes.  You should also cleanse your variables before putting them anywhere near your queries by calling mysql_real_escape_string() to prevent SQL Injections.

[noisyscanner]

Fixed it now!

Thanks for helping!