PHP Forms checkbox calculation

Posted by Sef on Stack Overflow See other posts from Stack Overflow or by Sef
Published on 2010-04-16T13:40:32Z Indexed on 2010/04/16 13:43 UTC
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Hello,

I am trying to perform some calculations with a form but every time i try to work with checkboxes it goes wrong.

The checkboxes are beign set on value 1 in the form itselff and are being checked if there checked or not.

$verdieping = isset($_POST["verdieping"]) ? $_POST["verdieping"] : 0;  
$telefoon = isset($_POST["telefoon"]) ? $_POST["telefoon"] : 0;  
$netwerk = isset($_POST["netwerk"]) ? $_POST["netwerk"] : 0; 

When i try to do calculations every works expect for the options with the checkboxes.
When both checkboxes (telefoon & netwerk) are selected the value should be 30.
If only one is selected the value should be 20.
But no mather what i have tried to write down it always give problem, and it always uses 20, never the value 30.

How do i solve this problem? Or suppose i am writing the syntax all wrong to lay conditions to a calculation? Any input appreciated.

$standnaam = $_SESSION["standnaam"];
$oppervlakte = $_SESSION["oppervlakte"];
$verdieping = $_SESSION["verdieping"];
$telefoon = $_SESSION["telefoon"];
$netwerk = $_SESSION["netwerk"];


if ($oppervlakte <= 10)
$tarief = 100;

if ($oppervlakte > 10 && $oppervlakte <= 20)
$tarief = 90;

if ($oppervlakte > 20)
$tarief = 80;


if($verdieping == 1)
{
$prijsVerdieping = $oppervlakte * 120;
}
else
{
$prijsVerdieping = 0;
}

if(($telefoon == 1) && ($netwerk == 1))
{
$prijsCom = 30; // never get this value, it always uses 20
}

if(($telefoon == 1) || ($netwerk == 1))
{
$prijsCom = 20;
}

$prijsOpp = $tarief * $oppervlakte; // works
$totalePrijs = $prijsOpp + $prijsVerdieping + $prijsCom; //prijsCom value is always wrong

Regards.

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