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Categorieën > PHP

informer - mysql_result werkt niet?

Nrzonline - 17/01/2006 21:16 (laatste wijziging 17/01/2006 23:50)
PHP ver gevorderde		dit is de header, die de $_SESSION['informer'] aanmaakt.. via $_['informer'].. In het begin gaf hij deze error niet, maar nu ik hem de meuk laat uploaden, en de oude accounts verwijderde begon hij deze error te geven: Warning: mysql_result(): supplied argument is not a valid MySQL result resource in C:\Program Files\YellowTip\Htdocs\warofthelords\header.php on line 14 Ik heb gezocht @ tutorials, maar daar kon ik niks vinden.. Ik heb gezocht @ php-net, maar daar werd ik niet veel wijzer uit =( Mocht je meerdere stukjes script nodig hebben, dan hoor ik dat graag.. ALVAST BEDANKT: header.php // oud -> removed.. zie volgende stukje code in post ±4

7 antwoorden

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nathanael - 17/01/2006 21:43 (laatste wijziging 17/01/2006 21:47)
HTML gevorderde		$CheckName_Result = mysql_result($CheckName_SQL,0) or (mysql_error()); lijk me dat het: $CheckName_Result = mysql_result($CheckName_SQL,0) or die(mysql_error()); moet zijn en: $CheckName_SQL = "SELECT user_name FROM site_users WHERE user_id = '" . $_SESSION['informer']; moet zijn: $CheckName_SQL = "SELECT user_name FROM site_users WHERE user_id = '" . $_SESSION['informer'] ."'; en als je user_id eeninteger is: $CheckName_SQL = "SELECT user_name FROM site_users WHERE user_id = '. $_SESSION['informer'].';

Rens - 17/01/2006 21:56

Crew algemeen

php code - Bekijk de code zonder highlighting - Klap code in

  $CheckName_Query = mysql_query($CheckName_SQL);

$CheckName_Query = mysql_query($CheckName_SQL);

Naar ==>

php code - Bekijk de code zonder highlighting - Klap code in

  $CheckName_Query = mysql_query($CheckName_SQL) or die (MySQL_Error());

$CheckName_Query = mysql_query($CheckName_SQL) or die (MySQL_Error());

php code - Bekijk de code zonder highlighting - Klap code in

    $CheckID_Query = mysql_query($CheckID_SQL);

$CheckID_Query = mysql_query($CheckID_SQL);

Naar ==>

php code - Bekijk de code zonder highlighting - Klap code in

    $CheckID_Query = mysql_query($CheckID_SQL) or die (MySQL_Error());

$CheckID_Query = mysql_query($CheckID_SQL) or die (MySQL_Error());

Gebruik error handling bij het uitvoeren van je queries!

Nrzonline - 17/01/2006 23:40 (laatste wijziging 18/01/2006 00:11)

PHP ver gevorderde

Het gaat nu al wat beter, maar wat me nu gebeurd snap ik dus echt
niet...

Error:

Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 7 in C:\Program Files\YellowTip\Htdocs\warofthelords\header.php on line 14

header.php

php code - Bekijk de code zonder highlighting - Klap code in

<?PHP

session_start();

require('includes/config.inc.php');
require('includes/functions.inc.php');
require('language/english.lang.php');

if (isset($_SESSION['informer']))
{
	$CheckName_SQL = "SELECT user_name FROM site_users WHERE user_id = '" . $_SESSION['informer'] . "'";
  	$CheckName_Query = mysql_query($CheckName_SQL) or die (mysql_error());
	$CheckName_Count = count($CheckName_Query) or die (mysql_error());
	$CheckName_Result = mysql_result($CheckName_Query,0) or die (mysql_error());
	
	if ($CheckName_Count == 1)
		$UserInformer = $CheckName_Result;
	else
		$UserInformer = "";
}
else
{
	if (!empty($_GET['informer']) && is_numeric($_GET['informer']))
	{
		$CheckID_SQL = "SELECT user_id FROM site_users WHERE user_id = '" . $_GET['informer'] . "'";
    		$CheckID_Query = mysql_query($CheckID_SQL) or die (mysql_error());
		$CheckID_Count = mysql_num_rows($CheckID_Query) or die (mysql_error());
		
		if ($CheckID_Count == 1)
		{
			$_SESSION['informer'] = $_GET['informer'];
			$CheckName_SQL = "SELECT user_name FROM site_users WHERE user_id = '" . $_SESSION['informer'] . "'";
			$CheckName_Query = mysql_query($CheckName_SQL) or die (mysql_error());
			$CheckName_Count = mysql_num_rows($CheckName_Query) or die (mysql_error());
			$CheckName_Result = mysql_result($CheckName_Query,0) or die (mysql_error()); 
			
			if ($CheckName_Count == 1)
				$UserInformer = $CheckName_Result;
		}
		else
			$UserInformer = "";
	}
	else
		$UserInformer = "";
}
?>

<?PHP
 
session_start();
 
require('includes/config.inc.php');
require('includes/functions.inc.php');
require('language/english.lang.php');
 
if (isset($_SESSION['informer']))
{
	$CheckName_SQL = "SELECT user_name FROM site_users WHERE user_id = '" . $_SESSION['informer'] . "'";
  	$CheckName_Query = mysql_query($CheckName_SQL) or die (mysql_error());
	$CheckName_Count = count($CheckName_Query) or die (mysql_error());
	$CheckName_Result = mysql_result($CheckName_Query,0) or die (mysql_error());
 
	if ($CheckName_Count == 1)
		$UserInformer = $CheckName_Result;
	else
		$UserInformer = "";
}
else
{
	if (!empty($_GET['informer']) && is_numeric($_GET['informer']))
	{
		$CheckID_SQL = "SELECT user_id FROM site_users WHERE user_id = '" . $_GET['informer'] . "'";
    		$CheckID_Query = mysql_query($CheckID_SQL) or die (mysql_error());
		$CheckID_Count = mysql_num_rows($CheckID_Query) or die (mysql_error());
 
		if ($CheckID_Count == 1)
		{
			$_SESSION['informer'] = $_GET['informer'];
			$CheckName_SQL = "SELECT user_name FROM site_users WHERE user_id = '" . $_SESSION['informer'] . "'";
			$CheckName_Query = mysql_query($CheckName_SQL) or die (mysql_error());
			$CheckName_Count = mysql_num_rows($CheckName_Query) or die (mysql_error());
			$CheckName_Result = mysql_result($CheckName_Query,0) or die (mysql_error()); 
 
			if ($CheckName_Count == 1)
				$UserInformer = $CheckName_Result;
		}
		else
			$UserInformer = "";
	}
	else
		$UserInformer = "";
}
?>

Ontani - 18/01/2006 00:19
-1		doe eens: echo $CheckName_SQL;

Thomas - 18/01/2006 07:40
Moderator		count(...) moet mysql_num_rows(...) zijn, lijkt me.

Voldemort - 18/01/2006 14:23 (laatste wijziging 18/01/2006 14:24)

PHP ver gevorderde

Maak van:

php code - Bekijk de code zonder highlighting - Klap code in

$CheckName_SQL = "SELECT user_name FROM site_users WHERE user_id = '" . $_SESSION['informer'] . "'";
$CheckName_Query = mysql_query($CheckName_SQL) or die (mysql_error());
$CheckName_Count = count($CheckName_Query) or die (mysql_error());
$CheckName_Result = mysql_result($CheckName_Query,0) or die (mysql_error());

$CheckName_SQL = "SELECT user_name FROM site_users WHERE user_id = '" . $_SESSION['informer'] . "'";
$CheckName_Query = mysql_query($CheckName_SQL) or die (mysql_error());
$CheckName_Count = count($CheckName_Query) or die (mysql_error());
$CheckName_Result = mysql_result($CheckName_Query,0) or die (mysql_error());

dat:

php code - Bekijk de code zonder highlighting - Klap code in

$CheckName_SQL = "SELECT COUNT(user_name) FROM site_users WHERE user_id = '" . $_SESSION['informer'] . "'";
$CheckName_Query = mysql_query($CheckName_SQL) or die (mysql_error());
$CheckName_Result = mysql_result($CheckName_Query,0);

$CheckName_SQL = "SELECT COUNT(user_name) FROM site_users WHERE user_id = '" . $_SESSION['informer'] . "'";
$CheckName_Query = mysql_query($CheckName_SQL) or die (mysql_error());
$CheckName_Result = mysql_result($CheckName_Query,0);

Error handling is enkel nodig bij mysql_query. Bij mysql_result niet. Verder kan je de count($CheckName_Query) rechtstreeks in je query steken. Nu kan je het aantal laten zien door $CheckName_Result.

Nrzonline - 18/01/2006 14:47 (laatste wijziging 18/01/2006 18:07)
PHP ver gevorderde		@Voldemort, ik zit nu even op school (vak technologie), dus ik kan het nu niet aanpassen... zogauw als ik thuis ben zou ik het direct proberen.. In iedergeval al hardstikke bedankt.. @rest, ook jullie allemaal bedankt =) Ok tried, werkt perfect! geen errors meer bij mijn regitratie-script TYVM ALL -> case closed =)

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