CODEWITHSUNDEEP
php
six.strings.and.a.bit
six.strings.and.a.bit

Reputation: 527

Why is the variable undefined in this code? - PHP

I have these codes:

testing.php

<html>
<head>
<script type="text/javascript">
        function copy_data(id){
            var a = document.getElementById(id).value;
            document.getElementById("copy_to").value=a;
        }
    </script>
</head>
<body>
<form action="testprocess.php" method="post">
<input type="text" name ="a" id="copy_from" onkeyup="copy_data('copy_from')"/>
<input type="text" value=0 name ="b" id="copy_to"/>
<input type="submit">
</form>
</body>
</html>

testprocess.php

<?php

    $test = $_POST['copy_to'];

    echo $test;
?>

I get an error saying that 'copy-to' is an undefined variable. Can you please tell me why? Thanks.

Upvotes: 0

Views: 95

Answers (5)

jcmeloni
jcmeloni

Reputation: 1234

$_POST will contain the values of your form fields based on the name attribute of the form elements.

<input type="text" name="copy_from"/> will become $_POST['copy_from']

and

<input type="text" name="copy_to"/> will become $_POST['copy_to']

You are using the value in the id attribute of the input (and spelling it inconsistently), so it is undefined to PHP.

Upvotes: 0

Clive
Clive

Reputation: 36955

$_POST values are passed through an element's name attribute rather than the ID. Try this:

<input type="text" value=0 name="copy_to" id="copy_to"/>

And make sure you use a an underscore in your PHP variable:

$test = $_POST['copy_to'];

Upvotes: 4

benhowdle89
benhowdle89

Reputation: 37504

Needs to be $_POST['a'] the id isn't submitted into the post array, it's the name attribute

Upvotes: 2

robert
robert

Reputation: 3615

It ought to be a _ instead of a -.

EDIT: Oh god, it's even worse. It ought to be $_POST['a'], because of the name-attribute. The name-attribute is used to specify the name/identifier under which a GET or POST parameter will be passed to the web application. The id-attribute is mostly used to identify HTML elements at the client, e.g. when doing some stuff with javascript.

Upvotes: -1

user319198
user319198

Reputation:

because you have no element with name copy_to in your form.

Try below :

<form action="testprocess.php" method="post">
<input type="text" name ="a" id="copy_from" onkeyup="copy_data('copy_from')"/>
<input type="text" value=0 name ="copy_to" id="copy_to"/>
<input type="submit">
</form>

Upvotes: 0

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