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Editorial Team
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Asked: 2026-06-03T18:40:38+00:00

this is the code i am working on (it is part of this tutorial

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this is the code i am working on (it is part of this tutorial http://www.homeandlearn.co.uk/php/php4p9.html )

this works fine

`

A BASIC HTML FORM

<?PHP

if (isset($_POST['Submit1'])) {
$username=$_POST['username'];
if($username=="ken"){
    print("you the man");
    }
    else {
        print("you are not supposed to be here");
        }
    }
    else{
        $username="";
        }

?>

</head>
<body>

<FORM NAME ="form1" METHOD ="post" ACTION = "basicForm.php">

username: <INPUT TYPE = "TEXT" VALUE ="<?PHP print $username;?>"NAME="username">

<INPUT TYPE = "Submit" Name = "Submit1" VALUE = "Login">

</FORM>
</body>
</html>`

But this does not

<html>
<head>
<title>A BASIC HTML FORM</title>

<?PHP


if (isset($_POST['Submit1'])) {

$username=$_POST['username'];
$nickname=$_POST['nickname'];
if($username=='ken'and$nickname=='hawk'){
    print("you the man");
    }
    else {
        print("you are not supposed to be here");
        }
    }
    else{
        $username=""and$nickname="";
        }

?>

</head>
<body>

<FORM NAME ="form1" METHOD ="post" ACTION = "testformken.php">

nickname: <input type="text" VALUE ="<?PHP print $nickname;?>" name="nickname" /><br />
username: <input type="text" VALUE ="<?PHP print $username;?>" name="username" />

<INPUT TYPE = "Submit" Name = "Submit1" VALUE = "Login">

</FORM>
</body>
</html>

I get this Notice: Undefined variable: nickname in C:\wamp\www\testformken.php on line 30 Call Stack #TimeMemoryFunctionLocation 10.1800367256{main}( )..\testformken.php:0 ” name=”nickname” />

I have messed with a few things and if i change 

nickname: <input type="text" VALUE ="<?PHP print $nickname;?>" name="nickname" /><br />
                      to
nickname: <input type="text" VALUE ="<?PHPprint$nickname;?>" name="nickname" /><br />

I do not get a the undifined variable but it does not print the nickname either
if i change the value to 

<?PHP print $username;?>

enter code here i do not get the undifined variable.

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1 Answer

  1. Editorial Team

    You’re simply always printing $nickname on your site but you only define it in the if-block (if (isset($_POST['Submit1'])) {) is entered.

    To avoid this, change your printing-part to:

    <?php print isset($nickname) ? $nickname : ''; ?>

    Note that I’m using the ternary operator (a shorthand for if-else) here which might be new for you if you’re a beginner.

    To avoid XSS-attacks you also shouldn’t output variables without escaping in real-life usage like Jack suggested (for a tutorial it’s ok to stay simple).

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