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Warning: mysql_num_row(): supplied argument is not a valid MySQL resul

 
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Warning: mysql_num_row(): supplied argument is not a valid MySQL resul
revense
8 Dec, 2006 - 08:34 PM
Post #1

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Joined: 8 Dec, 2006
Posts: 1


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Maybe this is a stupid question but I just started to learn PHP so, please help me to fix what wrong with my code

CODE
<?
//membuat session
//session_destroy();

// variabel yang diperlukan untuk akses database
$user = "root";
$pass = "admin";
$db = "pengaduan";
$server = "localhost";
// membuat koneksi
$koneksi = mysql_connect($server, $user, $pass);
// memeriksa koneksi

// membuka database
mysql_select_db($db);
// membuat query
$query = "SELECT * FROM LOGIN WHERE (USERNAME ='".$_POST['username']."') and (PASSWORDLOGIN='". md5($_POST['password'])."')";
// mengeksekusi query

$hasil = mysql_query($query);
$jumlahHasil = mysql_num_rows($hasil);

if($jumlahHasil <> 1)
    {
?>
    <script language=javascript>
    alert('Password atau Username salah !!')
    location.replace('index.php');
    </script>
<?
    }

else
    {
    $data = mysql_fetch_array($hasil);
    extract ($data);
    $passwordAsli = $_POST['password'];
    if($_POST['password'] == $passwordAsli)
        {
        session_start();
        $_SESSION['auth']="yes";
        $_SESSION['user']=$_POST['username'];
        $_SESSION['password']=$_POST['password'];
        $_SESSION['role']=$data[4];
        //echo $data[4];
        //echo $_SESSION['role'];
        //echo "<head><meta http-equiv='refresh' content='0;url=\"menu.php\"'></head>";
        ?>
        <script language=javascript>
        alert('Login Berhasil !!')
        location.replace('menu.php');
        </script>
        <?
        }
    else
        {
        echo("Password untuk user $USERNAME salah<br>");
        }
    }
?>


When I try this script, I get this warning...

Warning:mysql_num_rows():supplied argument is not a valid Mysql result resource

will you help me what wrong
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Amadeus
RE: Warning: Mysql_num_row(): Supplied Argument Is Not A Valid MySQL Resul
9 Dec, 2006 - 05:30 AM
Post #2

g++ -o drink whiskey.cpp
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Have you checked to see if anything is being returned from your query? Or that it has executed properly? mysql_num_rows will only take a valid resource as a parameter, and if the query failed, then no valid resource will be returned.

Try this:
CODE

$hasil = mysql_query($query);
if (!$hasil) {
   echo "Query Error: " . mysql_error() . "\n";
}
else
{
   echo "Query is fine\n";
}
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manilodisan
RE: Warning: Mysql_num_row(): Supplied Argument Is Not A Valid MySQL Resul
10 Dec, 2006 - 01:10 PM
Post #3

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Joined: 10 Dec, 2006
Posts: 2


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try replacing your query line with this ones:


CODE
$query = "SELECT * FROM LOGIN WHERE (USERNAME ='".$_POST['username']."') and (PASSWORDLOGIN='". md5($_POST['password'])."')";
$hasil = mysql_query($query) or die(mysql_error());
$row_hasil = mysql_fetch_assoc($hasil);
$jumlahHasil = mysql_num_rows($hasil);
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tork
RE: Warning: Mysql_num_row(): Supplied Argument Is Not A Valid MySQL Resul
14 Dec, 2006 - 03:09 PM
Post #4

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Joined: 14 Dec, 2006
Posts: 20


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the error that you stated: Warning:mysql_num_rows():supplied argument is not a valid Mysql result resource

In my time so far in PHP, this error appears when your query results in nothing. Make sure you know that there will be a result before you do mysql_num_rows or carry out:

CODE
if ($result){

$count = mysql_num_rows($result)

}


i do this as a safety precaution.

QUOTE(revense @ 8 Dec, 2006 - 09:34 PM) *

Maybe this is a stupid question but I just started to learn PHP so, please help me to fix what wrong with my code

code


<?
//membuat session
//session_destroy();

// variabel yang diperlukan untuk akses database
$user = "root";
$pass = "admin";
$db = "pengaduan";
$server = "localhost";
// membuat koneksi
$koneksi = mysql_connect($server, $user, $pass);
// memeriksa koneksi

// membuka database
mysql_select_db($db);
// membuat query
$query = "SELECT * FROM LOGIN WHERE (USERNAME ='".$_POST['username']."') and (PASSWORDLOGIN='". md5($_POST['password'])."')";
// mengeksekusi query

$hasil = mysql_query($query);
$jumlahHasil = mysql_num_rows($hasil);

if($jumlahHasil <> 1)
{
?>
<script language=javascript>
alert('Password atau Username salah !!')
location.replace('index.php');
</script>
<?
}

else
{
$data = mysql_fetch_array($hasil);
extract ($data);
$passwordAsli = $_POST['password'];
if($_POST['password'] == $passwordAsli)
{
session_start();
$_SESSION['auth']="yes";
$_SESSION['user']=$_POST['username'];
$_SESSION['password']=$_POST['password'];
$_SESSION['role']=$data[4];
//echo $data[4];
//echo $_SESSION['role'];
//echo "<head><meta http-equiv='refresh' content='0;url=\"menu.php\"'></head>";
?>
<script language=javascript>
alert('Login Berhasil !!')
location.replace('menu.php');
</script>
<?
}
else
{
echo("Password untuk user $USERNAME salah<br>");
}
}
?>
/code

When I try this script, I get this warning...

Warning:mysql_num_rows():supplied argument is not a valid Mysql result resource

will you help me what wrong
User is offlineProfile CardPM
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Chemo
RE: Warning: Mysql_num_row(): Supplied Argument Is Not A Valid MySQL Resul
20 Dec, 2006 - 05:40 PM
Post #5

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Joined: 17 Dec, 2006
Posts: 2


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Try this:
CODE

<?php
/**
* Simple refactor for "revense" Dreamincode.net forum member
*
* Login script
*
* @license http://opensource.org/licenses/gpl-license.php GNU Public License
* @version 1.0
* @link http://www.oscommerce-freelancers.com/ osCommerce-Freelancers
* @copyright Copyright 2006, Bobby Easland
* @author Bobby Easland
* @filesource
*/

    /**
     * Define the DB connection params
     */
    define('DB_USER', 'root');
    define('DB_PASS', 'admin');
    define('DB_NAME', 'pengaduan');
    define('DB_HOST', 'localhost');
    
    /**
     * Connect to the DB
     */
    $dbLink = mysql_connect(DB_SERVER, DB_USER, DB_PASS);
    
    /**
     * Basic connection sanity check
     */
    if (!$dbLink) {
      die( 'Could not connect to database -> ' . mysql_errno() . ': ' . mysql_error() );
    }

    /**
     * Select the DB
     */
    $dbSelect = mysql_select_db(DB_NAME, $dbLink);
    
    /**
     * Basic DB select sanity check
     */
    if (!$dbSelect) {
         die( 'Cannot select DB -> ' . mysql_errno() . ': ' . mysql_error() );
    }    
    
    /**
     * Initialize the SQL
     */
     $sql = "SELECT *
             FROM LOGIN
                     WHERE USERNAME = '".mysql_real_escape_string($_POST['username'])."'
                         AND PASSWORDLOGIN = '". md5(mysql_real_escape_string($_POST['password']))."'
                     LIMIT 1";
    
    /**
     * Execute query
     */
    $query = mysql_query($sql, $dbLink);
    
    /**
     * Basic sanity check
     */
    if (!$query){
        die( 'Could not execute query -> ' . mysql_errno() . ': ' . mysql_error()  );
    }
    
    /**
     * Get the number of rows...should be 1 per the SQL limit
     */
    $rowsReturned = mysql_num_rows($query);

    if($rowsReturned !== 1) {
    ?>
        <script language=javascript>
        alert('Password atau Username salah !!')
        location.replace('index.php');
        </script>
    <?
    } else {
        $data = mysql_fetch_assoc($query);
        extract($data);
        $passwordAsli = $_POST['password'];
        if( $_POST['password'] == $passwordAsli ){
            session_start();
            $_SESSION['auth'] = "yes";
            $_SESSION['user'] = $_POST['username'];
            $_SESSION['password'] = $_POST['password'];
            $_SESSION['role'] = $data[4];
        ?>
        <script language=javascript>
        alert('Login Berhasil !!')
        location.replace('menu.php');
        </script>
        <?
        } else {
            echo("Password untuk user $USERNAME salah<br>");
        }
    }
?>
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AdamG
RE: Warning: Mysql_num_row(): Supplied Argument Is Not A Valid MySQL Resul
21 Dec, 2006 - 07:57 AM
Post #6

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Posts: 94



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As stated above, I can almost guarantee the problem is your query. Adding "or die(mysql_error())" will display the SQL error received and help troubleshoot your problem.

--Adam
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maxbox
RE: Warning: Mysql_num_row(): Supplied Argument Is Not A Valid MySQL Resul
11 Dec, 2008 - 05:36 PM
Post #7

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Joined: 7 Nov, 2008
Posts: 3
You can try with something like this


CODE
if (mysql_num_rows($query) <> 1 )



{
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