[Resolved] mysql_fetch_array incorrect

[demosthenes705]

I have the following code

<?php
include ("db.php");

$query = mysql_query("SELECT * FROM `uniquehits` WHERE `ip` = '.$_SERVER['REMOTE_ADDR'].'");

if( !mysql_fetch_array( $query ) ) { 
mysql_query( "INSERT INTO `uniquehits` VALUES( null, '".$_SERVER['REMOTE_ADDR']."' )" ); 
}; 

$result = ("SELECT `id` FROM `uniquehits` WHERE 1 ORDER BY `id` DESC LIMIT 1"); 
$query = mysql_fetch_array( mysql_query( $query, MYSQL_ASSOC ) ) ; 
$query = $query['id']; 

echo "We have received $query total hits since 3-7-2005."; ?>

And when I run it i get the following error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/vhosts/scn.xoopiter.com/member/admin/stats.php on line 6

I got that code from a site. I cannot find it anymore. So I am wondering what I need to fix.

[devinemke]

replace this:

$query = mysql_query("SELECT * FROM `uniquehits` WHERE `ip` = '.$_SERVER['REMOTE_ADDR'].'");

with this:

$query = mysql_query("SELECT * FROM uniquehits WHERE ip = '" . $_SERVER['REMOTE_ADDR'] . "'") or exit(mysql_error());

[demosthenes705]

well i gave up because it wasn't working

[devinemke]

Originally posted by demosthenes705
well i gave up because it wasn't working

that's the spirit! right on!

[demosthenes705]

the reason i did give up was because it wasn't adding it once and only once. it was adding it everytime i went there