Help: Simple mysql_fetch_array error...

Discussion in 'PHP' started by Skillman13, Nov 23, 2009.

0
  1. #1
    I can't get this to work...

    $query = mysql_query("SELECT * FROM `Queue` WHERE `GameId` <= 5 AND `Username` != '$username1' AND `GunTactics` = '9mm Uzi'");
    $result = mysql_query($query);
    $row = mysql_fetch_array($result);
    $username2 = $row['Username'];

    It keeps giving me this error...

    Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/mountgam/public_html/zombie/killhousecron.php on line 353


    ($username1 = 'Test' by the way)
    Can anyone fix this?

    Thanks alot,

    James
     
    Skillman13, Nov 23, 2009 IP
  2. mastermunj

    mastermunj Well-Known Member

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    #2
    change following lines

    
$query = mysql_query("SELECT * FROM `Queue` WHERE `GameId` <= 5 AND `Username` != '$username1' AND `GunTactics` = '9mm Uzi'");
$result = mysql_query($query);
    PHP:
    TO

    
$result = mysql_query("SELECT * FROM `Queue` WHERE `GameId` <= 5 AND `Username` != '$username1' AND `GunTactics` = '9mm Uzi'");
    PHP:
     
    mastermunj, Nov 23, 2009 IP
  3. ColdMoon

    ColdMoon Peon

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    #3 
    $query = mysql_query("SELECT * FROM Queue WHERE GameId<=5 AND Username!='$username1' AND GunTactics='9mm Uzi'") or die(mysql_error());

while ($data = mysql_fetch_array($query)) {
    echo "Username: " . $data['Username'] . "<br />";
}
    PHP:
     
    ColdMoon, Nov 23, 2009 IP