PHP Forum

**tunjiaodu** · 03-21-2008, 06:42 PM

I am trying to display the number of acticve users in my table using the code below. but i geet this error

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/globalpe/public_html/tunji2.php on line 13.
<?php
$dbhost = "";
$dbuser = "";
$dbpass = "";
$dbname = "";

$link = mysql_connect($dbhost, $dbuser, $dbpass)
or die(mysql_error());
mysql_select_db($dbname)
or die(mysql_error());

$sql = 'SELECT COUNT(`ID`) FROM `Profiles` WHERE `Status` = \'Active\'';
$results = mysql_result($sql, "0")
or die(mysql_error());
print($results);
?>

**NullPointer** · 03-21-2008, 11:51 PM

mysql_result does not take a string as input, it takes the result of a mysql query. Your code is fine more or less but you simply forgot a step. Try this:

PHP Code:


			
<?php 

$dbhost = ""; 

$dbuser = ""; 

$dbpass = ""; 

$dbname = ""; 



$link = mysql_connect($dbhost, $dbuser, $dbpass) 

or die(mysql_error()); 

mysql_select_db($dbname) 

or die(mysql_error()); 



$sql = 'SELECT COUNT(`ID`) FROM `Profiles` WHERE `Status` = \'Active\'';

$result = mysql_query($sql); 

$results = mysql_result($result, "0") 

or die(mysql_error()); 

print($results); 

?>