$_POST ['variable'], variable not recognized ( "Undefined index" )

Question

I am trying to learn PHP (just a tiny bit for a tiny project). I am trying to follow this tutorial, but when I do, it does not work. First of all I coppied the exact code there and it still did not work, but here is some of my code:

    <?php
         if($_POST['projectSubmit'] == "Submit") 
        {
            $toApprove= $_POST['projectName'];
        }
    ?>


    <form action ="getData" method="post" >
        <input type="text" name="projectName">
        <input type="submit" name="projectSubmit" value="Submit">
    </form>

Yet I get an error:

A PHP Error was encountered

Severity: Notice

Message: Undefined index: projectSubmit

Filename: views/ViewProjectApproval.php

Line Number: 13

What am I doing wrong ?

BTW: Is this a correct way of transmitting data back to my controller ? (model view controller with code igniter)

you could turn these notices off http://stackoverflow.com/questions/2867057/how-do-i-turn-off-php-notices BUT I suggest keeping them on for now. — Waygood, Apr 24 '13 at 15:43
@Waygood So this is a non fatal error then ? Mind explaining what caused it though ? — Kalec, Apr 24 '13 at 15:45
chandresh_cool already answered it. The explanation is that you are trying to access an element of $_POST that hasn't been created yet. (it is AFTER you submit the form). And BTW this is not MVC. — Waygood, Apr 24 '13 at 15:46
Possible duplicate of [PHP: “Notice: Undefined variable” and “Notice: Undefined index”](http://stackoverflow.com/q/4261133/1409082) — Jocelyn, Apr 24 '13 at 15:50

score 3 · Answer 1 · answered Apr 24 '13 at 15:41

3

Use:

if (isset($_POST['projectSubmit'])) { /*...*/ }

This checks if the parameter is sent. (here: when the form was submitted)

answered Apr 24 '13 at 15:41

bwoebi

23,637
5
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79

score 2 · Accepted Answer · edited Apr 24 '13 at 15:50

2

Use this

if (isset($_POST['projectSubmit']) && $_POST['projectSubmit'] == "Submit")

edited Apr 24 '13 at 15:50

bwoebi

23,637
5
58
79

answered Apr 24 '13 at 15:41

chandresh_cool

11,753
3
30
45

Do you have any idea what exactly was wrong with mine ? I don't wanna follow tutorials from a site with bad code. – Kalec Apr 24 '13 at 15:45
2

You were directly checking the value of $_POST['projectSubmit'] you shoud first check is that post variable is available or not using isset because POST value only comes when you submit the form using button. – chandresh_cool Apr 24 '13 at 15:48

$_POST ['variable'], variable not recognized ( "Undefined index" )

2 Answers2