PHP-script errors..

DeniseSamuela · 10-Jun-2012, 10:55 AM #1

Hi,

I've tried to test my php-gamescript in UsbWebserver 7.0.

PHP Code:

 
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">

 
<html>

<head>

<title>Untitled</title>

</head>

<body>

<?php

function letters_only($string)  

{ 

$eregi = eregi_replace("([A-Z]+)","",$string); 

if(empty($eregi)) { 

return true; 

} 

return false; 

} 

if(!$POST["submit"])  

    {  

    $woorden = array("geweldig", "condensatiepolymerisatie", "additiepolymerisatie");   

    $random= rand(0,count($woorden)-1);  

    $woord=$woorden[$random];  

    $sessie["woord"]=$woord;  

    $teller=7;   

    for ($x=0;$x<strlen($woord);$x++){  

    $verborgen.="-";}  

    }   

else  

    {  

    $letter=$POST["letter"];  

    $woord=$_sessie["woord"];  

    $hiddenword=$POST["hidden"];  

    $gebruikt=$POST["gebruikt"];  

    $teller=$POST["teller"];  

    $gebruikt.=$letter;  

    $verborgen="";  

    $juist=false;  

    }   

 
 
if(substr($woord,$x,1)==$letter) 

{ 

$verborgen.=$letter; 

$juist=true; 

} 

 
else 

{ 

$verborgen.=substr($verborgenwoord,$x,1); 

} 

  

 
if(!$juist) 

{  

$teller--; 

} 

 
if ($woord==$verborgen) 

{ 

echo "Gefeliciteerd u heeft het woord geraden<br>"; 

echo "Het woord was: ".$woord; 

$end=true; 

if ($teller==0) 

echo "U heeft 7 keer geraden, u bent af";  

echo "<br> Het woord was: ".$woord; 

} 

 
if($end=true); 

 
 
if (!$end) 

{ 

echo "Voorlopig heeft u dit geraden:<br>"; 

echo $verborgen; 

echo "<br>U heeft nog ".$teller." levens<br>"; 

if($gebruikt) 

echo "Deze letters heeft u al gebruikt:" .strtoupper($gebruikt); 

}  

?>

</body>

</html>

The test run says the following:
Notice: Undefined variable: POST in F:\UsbWebserver\Root\game.php on line 17

Notice: Undefined variable: verborgen in F:\UsbWebserver\Root\game.php on line 25

Notice: Undefined variable: letter in F:\UsbWebserver\Root\game.php on line 40

Notice: Undefined variable: letter in F:\UsbWebserver\Root\game.php on line 42

I really don't understand why the test run says this, because, for example, my $POST is defined in (!$POST["submit"]) , right?

Can anyone help me with improving my script? I really want my script to be 100% working

Kind regards,

Denise

allnodcoms · 11-Jun-2012, 03:57 AM #2

That should be $_POST["submit"] (note underscore)...

That will probably fix your script

Danny

dukevyner · 11-Jun-2012, 09:28 PM #3

Quote:

Originally Posted by allnodcoms

That should be $_POST["submit"] (note underscore)...

That will probably fix your script

Danny

Yup, he's right, if this solves the problem be sure to click the "mark solved" button at the top of the page

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