Solved

Transer parameters from one php page to onother

Posted on 2013-06-12
3
172 Views
Last Modified: 2013-06-13
Hi!

Have this php-page

<?php // RAY_EE_login.php
require_once('RAY_EE_config.php');
$table1 = 'Kontroller';
// ACCESS TO THIS PAGE IS TESTED BUT NOT CONTROLLED
if ($uid = access_control(TRUE))
{
   $brukerID = brukerident('xxx');
}
else
{
        header("location: login.php");
	    exit;
}

?>
<!DOCTYPE html>

<html>
<head>
<meta charset="utf-8" />
<title>Test system</title>

</head>

<body>
<link rel="stylesheet" href="default.css">
<meta name="viewport" content="width=device-width, initial-scale=1, maximum-scale=1">


<table summary="Services, or Links box template" class="servicesT" cellspacing="1">
<tr><td colspan="2" class="servHd">Min kontroller</td></tr>
<br><br>
<?php
$result3 = mysql_query("SELECT Navn FROM {$table1} WHERE Bruker_id='{$brukerID}'") or die(mysql_error());
while ($row3 = mysql_fetch_array($result3)){
    $navn = $row3['Navn'];
	$lokk = $row3['Lokasjons_id'];
?>
<tr>
	<td class="servBodL_medborder" ><?php echo $navn; ?></td>
	<td class="servBodL_medborder"><a href="http://www.softkey.no/webtest/side3.php?Navn=<?php echo $navn; ?>&lokk=<?php echo $lokk; ?>&IDENT=<?php echo $brukerID; ?>"><input type="submit" id="GULKN" value=">"></a></td>
</tr>
<?php
}

?>

</table>
<br>
<table summary="Services, or Links box template" class="servicesT" cellspacing="0">
<tr><td colspan="2" class="servHd">Mine åpne kontroller</td></tr>
<?php

$result3 = mysql_query("SELECT Navn FROM {$table1} WHERE status=1 and Bruker_id='{$brukerID}'") or die(mysql_error());
while ($row3 = mysql_fetch_array($result3)){
    $navn = $row3['Navn'];
?>
<tr>
	<td class="servBodL_medborder" ><?php echo $navn; ?></td>
	<td class="servBodL_medborder"><a <input type="submit" id="GULKN" value=">" href="http://www.test.com/newpage.php?ID=20?IDENT=40"></a></td>
</tr>
<?php
}

?>
</table>
<br><br>
<Form method="post" action="logut.php">
<fieldset id="actions">
	<input type="submit" id="LOGIN" value="Logg ut">
</fieldset>
</Form>

</body>
</html>

Open in new window


My probles is that variable $lokk is emty, and i know that in the database
it contains data...

Why is the variable emty ?
0
Comment
Question by:team2005
3 Comments
 
LVL 11

Accepted Solution

by:
Amar Bardoliwala earned 500 total points
ID: 39240637
Hello  team2005,

It seems that in your 'SELECT' query you are only fetching column 'Navn'. Please change you query as following.

$result3 = mysql_query("SELECT Navn, Lokasjons_id FROM {$table1} WHERE Bruker_id='{$brukerID}'") or die(mysql_error());

Hope, this will help you.

Thank you.

Amar.
0
 
LVL 9

Expert Comment

by:TvMpt
ID: 39240642
To help you debug just make var_dump($result3) after mysql query.

Could be a multidemensional array or just empty too
0
 
LVL 2

Author Closing Comment

by:team2005
ID: 39243772
Thanks for help
0

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