what does this mean - Webmaster Forum - Web Design, Programming and SEO forums

03-01-2007, 03:06 AM

Taylor Hewitt

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what does this mean

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/teeyaych/public_html/emogroups/test/admin/index.php on line 147

Warning: Cannot modify header information - headers already sent by (output started at /home/teeyaych/public_html/emogroups/test/admin/index.php:147) in /home/teeyaych/public_html/emogroups/test/admin/index.php on line 156

Warning: Cannot modify header information - headers already sent by (output started at /home/teeyaych/public_html/emogroups/test/admin/index.php:147) in /home/teeyaych/public_html/emogroups/test/admin/index.php on line 158

03-01-2007, 03:13 AM

aLx

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Post up your script please.. or that page

03-01-2007, 03:16 AM

Taylor Hewitt

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Code:

function Logout($admin) {

    $cookie = explode("|", base64_decode($admin));
    $result = mysql_query("SELECT password FROM ".$prefix."_admin WHERE admin_name='$cookie[1]'");
    $row = mysql_fetch_array($result);
    $pass = $row['password'];
    if ($cookie[2] == $pass && $pass != "") {
	return $cookie;
    } else {
	unset($admin);
	unset($cookie);
    }
    
    setcookie("admin", false);
    $admin = "";
    header("Location: index.php");
    
}

03-02-2007, 02:10 PM

bluesaga

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The first error message means the mysql query you are trying to run is errorneous, change it from:

PHP Code:


			
    $result = mysql_query("SELECT password FROM ".$prefix."_admin WHERE admin_name='$cookie[1]'");

to:

PHP Code:


			
$result = mysql_query("SELECT password FROM ".$prefix."_admin WHERE admin_name='$cookie[1]'")

   or die( mysql_error() );

This should give you a better error message telling you exactly what is wrong with the query. Could be anything, but its bugged

the latter is regarding setting your cookies and headers, because when creating a cookie or setting a location header to move to a different place, NO output should be produced before these functions are called, including a space at the top of your document so watch out for these

03-03-2007, 02:23 AM

WarlordTony

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I think the query is wrong because of $prefix var - it is not declared within the function and tries to query a non existent table "_admin".

03-03-2007, 02:39 AM

Andrew R

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WarlordTony, I'm sure he declares it in a config.php file.

PHP Code:


			
$query = sprintf("SELECT password FROM %s_admin WHERE admin_name='%s'", $prefix, $cookie[1]);

$result = mysql_query($query);

Try that.

03-03-2007, 02:44 AM

Jeff Andersen

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And as for the second/third error, you cannot set a cookie after any browser code (such as including a header or anything to do with the client side page) has been output.

03-03-2007, 02:16 PM

bluesaga

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Originally Posted by Andrew R

WarlordTony, I'm sure he declares it in a config.php file.

PHP Code:


			
$query = sprintf("SELECT password FROM %s_admin WHERE admin_name='%s'", $prefix, $cookie[1]);

$result = mysql_query($query);

Try that.

Warlord tony is correct, how can he declare a variable and call it WITHIN a function without using globals?

Id imagine this is the source of his errors aswell.