mysql_result error

[Canman2005]

Hi all

 

I have the following code

 

<?php
$db_name ="***";
$server = "localhost";
$dbusername = "***";
$dbpassword = "***";

$connection = mysql_connect($server, $dbusername, $dbpassword) or die(mysql_error());
$db = mysql_select_db($db_name,$connection)or die(mysql_error());

$sql = "SELECT * FROM countries ORDER BY description ASC";
$query = @mysql_query($sql,$connection) or die(mysql_error());

while ($row = mysql_fetch_array($query))
{
$total = "SELECT COUNT(*) as Num FROM users WHERE `email` NOT LIKE '%@googlemail.com%' AND country_code = ".$row['code']."";
$results = mysql_result(mysql_query($total),0); 
?>
<?php print $row['description']; ?><br>
<?php print $results; ?><br><br>
<?php
}
?>

 

But when I run the page, I get the error

 

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/public_html/test.php

 

Does anyone know why?

 

I've checked and checked again, but cannot seem to find a answer

 

Any help would be great

 

Thanks

 

Dave

[jitesh]

Try this

 

<?php
$total = "SELECT COUNT(*) as Num FROM users WHERE `email` NOT LIKE '%@googlemail.com%' AND country_code = ".$row['code']."";
$results = mysql_query($total,$connection);
$Num = mysql_result($results,0,'Num');
?>

[xenophobia]

$results = mysql_result(mysql_query($total),0);

One more argument, the field name of your table.

 

mysql_result(mysql_query($total), 0, "field_1");

[Canman2005]

thanks, works fine now

[hvle]

that's because your query $total is not in correct syntax:

 

fix:

$total = "SELECT COUNT(*) as Num FROM users WHERE `email` NOT LIKE '%@googlemail.com%' AND country_code = '{$row['code']}'";