mysql_num_rows() and mysql_fetch_array() errors

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July 21st, 2004, 01:35 AM

asiansINC.com

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mysql_num_rows() and mysql_fetch_array() errors

I've googled this site extensively, but can't figure out what's wrong with the script that I'm trying to run.

Here's the code:

PHP Code:


		
			
<?
session_start();
session_destroy();
$who=$_GET['from'];
$ref=$_SERVER['HTTP_REFERER'];
unset($HTTP_SESSION_VARS["sponsor"]);
unset($HTTP_SESSION_VARS["random"]);
setcookie("sponsor_is", "", time()-3600);
setcookie("sponsor_is", $who, time()+24*60*60*365);
require_once("db_connect.php");
$s=mysql_query("SELECT * FROM users WHERE username='.mysql_real_escape_string(strtolower($wh  o)).' and status=1");
//if no valid sponsor entered, let's just pay the admin
if(mysql_num_rows($s)<1){
 $s=mysql_query("SELECT * FROM users WHERE id=1");
 }
@session_register("sponsor");
$HTTP_SESSION_VARS["sponsor"]=mysql_fetch_array($s);
$r=mysql_query("INSERT INTO 'hits' ('username' , 'refer' , 'ip' , 'date' )
VALUES ( '".$who."', '$ref', '$REMOTE_ADDR',now())");
echo "<meta http-equiv='Refresh' content='0; url=index.php'>";
?>

This is the error that I get:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home2/random/public_html/promo.php on line 13

So I tried commenting out lines 13-15 if(mysql_num_rows($s)<1){$s=mysql_query("SELECT * FROM users WHERE id=1");}

But then running the script gave me this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home2/random/public_html/promo.php on line 17

I've tried adding or die in various places, but no errors showed up.

Any help is greatly appreciated.

July 21st, 2004, 07:36 AM

dhouston

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The call to mysql_fetch_array() is failing because you commented out the mysql_query(), which fails because you don't have a valid connection to the database. Paste in your db_connect.php (change username and password values) and let's have a look at that.

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July 21st, 2004, 08:26 PM

asiansINC.com

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This is db_connect.php:

PHP Code:


		
			
<? 
/*This is the mysql connection parameters.
For maximum security,you can move this file outside the 
web tree .
In this case,please edit the path from the line 100 of config.php.*/
$dbhost = 'localhost';
$database_name = 'database_name';
$dbusername = 'database_username';
$dbpasswd = 'database_password';
$connection = mysql_pconnect("$dbhost","$dbusername","$dbpasswd") 
 or die ("Couldn't connect to server.");
 $db = mysql_select_db("$database_name", $connection)
 or die("Couldn't select database.");
 ?>

July 22nd, 2004, 08:36 AM

dhouston

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Try prepending "@" to all of your mysql functions. The errors you're getting are warnings and so this may just be a matter of your log level being set too high. Sticking "@" at the beginning of the functions should suppress the errors, while having die statements will print out any real problems. I'd recommend adding the output of mysql_error() to your die statements during development.

July 25th, 2004, 01:17 AM

asiansINC.com

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Thank you, dhouston.

Can you give me an example of where I should be prepending "@"?

July 26th, 2004, 07:54 AM

dhouston

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Code:

$connection = @mysql_pconnect("$dbhost","$dbusername","$dbpasswd") or die ("Couldn't connect to server.");

July 29th, 2004, 07:33 PM

tim_ver

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Error ?

I get some what of the same error mesg.

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/main/public_html/Job8/meta.inc on line 7

Is this the same?

Thanks

July 29th, 2004, 10:07 PM

dhouston

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Add the @ sign to that function as well, and to any others that generate the error.

July 29th, 2004, 10:47 PM

tim_ver

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Changed it..

Ok for this line 7 is:

$meta_row = mysql_fetch_array($meta_return);

I changed it to: @meta_row = mysql_fetch_array($meta_return);

I get a parse error.

I tried

$meta_row = @mysql_fetch_array($meta_return);

And it went away. Is this ok or just a band aid?

Thanks

#10

July 30th, 2004, 08:16 AM

dhouston

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See response at http://forums.devarticles.com/t9344/s.html. In short, it's somewhere between a band-aid and a fix.

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