Undefined variable error when the variable name is used in form filed

Question

while executing my code some times a warning message comes like use of undefined variable even if the variable name is being used properly in the form field.

<form method="post" id="" action"">
<input type="text" name="name" />
<input type="submit" name="submit" value="submit">
</form>

$name=$_POST['name'];

even if the name is used in the form field warning message comes and i had to put @ in front of it

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

@$name=$_POST['name'];
}

why is is so can anyone help me out with it

Who the hell is a Down Voter Has down voted every answer without a single comment about it :( — GajendraSinghParihar
– GajendraSinghParihar, Commented Feb 27, 2014 at 7:22
Sidenote: You're missing an = sign in action"" so that won't work. — Funk Forty Niner
– Funk Forty Niner, Commented Feb 27, 2014 at 15:31

Rakesh Sharma · Accepted Answer · 2014-02-27 07:10:23Z

2

the reason was you have not initialized $name variable and also $_POST['name'] not found until your form not submitted

so try to initialize variable first like

$name = '';

or

$name= (isset($_POST['name']) ? $_POST['name'] : '');

or

if(isset($_POST['name']))
        $name=$_POST['name'];

answered Feb 27, 2014 at 7:10

Rakesh Sharma

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Shankar Narayana Damodaran · Accepted Answer · 2014-02-27 07:08:10Z

1

You are not at all submitting your <form> using a submit button. and the action attribute has no =

Should be like ...

Your HTML form

<form method="post" action="test.php">
<input type="text" name="name" />
<input type="submit" name="submit" />
</form>

test.php

if(isset($_POST['name']))
{
 echo $_POST['name'];
}

answered Feb 27, 2014 at 7:08

Shankar Narayana Damodaran

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Comments

GajendraSinghParihar · Accepted Answer · 2014-02-27 07:12:41Z

0

USE $name = (isset($_POST["name"])?$_POST["name"]:"");

answered Feb 27, 2014 at 7:12

GajendraSinghParihar

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