How to fix 'Notice: Undefined index:' in PHP form action

Question

I received the following error message when I tried to submit the content to my form. How may I fix it?

Notice: Undefined index: filename in D:\wamp\www\update.php on line 4

Example Update.php code:

<?php

    $index = 1;
    $filename = $_POST['filename'];

    echo $filename;
?>

And $_POST['filename'] comes from another page:

<?php
    $db = substr($string[0],14) . "_" . substr($string[1],14) . "_db.txt";
?>

<input type="hidden" name="filename" value="<?php echo $db; ?>">

D:\wamp\www\update.php, Are you accessing page this way or it just a path to the file? Have you tried it on localhost? — Jai
– Jai, Commented Dec 31, 2012 at 5:56
Notice: Undefined index: filename in D:\wamp\www\update.php on line 4, obviously he's accessing it through localhost or he wouldn't get that error. — David Harris
– David Harris, Commented Dec 31, 2012 at 5:59
We can't help you anymore OP, we will get no where just guessing. Please post the entire code. — David Harris
– David Harris, Commented Dec 31, 2012 at 6:01

Sir · Accepted Answer · 2014-10-29 19:52:19Z

48

Assuming you only copy/pasted the relevant code and your form includes <form method="POST">

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

If _POST is not set the filename variable won't be either in the above example.

An alternative way:

$filename = false;
if(isset($_POST['filename'])){
    $filename = $_POST['filename'];
 } 
    echo $filename; //guarenteed to be set so isset not needed

In this example filename is set regardless of the situation with _POST. This should demonstrate the use of isset nicely.

More information here: http://php.net/manual/en/function.isset.php

edited Oct 29, 2014 at 19:52

answered Dec 31, 2012 at 5:48

Sir

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7 Comments

Sir Over a year ago

It depends at what stage of the script he wants to echo it, he can do that also if he wishes.. but then why assign it to $filename at all :P It was really just to demonstrate the use of isset

Sir Over a year ago

I personally would just do echo $_POST['filename'];

David Harris Over a year ago

You're missing the point though. It doesn't matter where you use isset(), it can always be the same as isset(GET FILENAME) and isset(filename)

Sir Over a year ago

Its a shame i expected his form to be complete not as it was shown, thought he was just showing a snippet ! Good spot with that!

David Harris Over a year ago

Why is this the best answer? It's wrong in almost every aspect.

|

Jim Bergman · Accepted Answer · 2017-03-26 21:18:50Z

7

if(isset($_POST['form_field_name'])) {
    $variable_name = $_POST['form_field_name'];
}

edited Mar 26, 2017 at 21:18

Jim Bergman

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answered Mar 26, 2017 at 19:39

Rabby shah

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Comments

Ryan Wildry · Accepted Answer · 2018-05-29 16:38:19Z

5

Change $_POST to $_FILES and make sure your enctype is "multipart/form-data"

Is your input field actually in a form?

<form method="POST" action="update.php">
    <input type="hidden" name="filename" value="test" />
</form>

edited May 29, 2018 at 16:38

Ryan Wildry

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answered Dec 31, 2012 at 5:45

David Harris

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6 Comments

Ting Ping Over a year ago

Yes, it's enclosed within the form field. I didn't include above, that's all.

ravi404 Over a year ago

@TingPing I dont find any other issues with your code. Is that your complete code

David Harris Over a year ago

That's a very key thing to include in your post, Ting. What happens when you var_dump($_POST) ?

ravi404 Over a year ago

var_export($_REQUEST) with var_dump($_POST) could be checked.

David Harris Over a year ago

var_export isn't really needed, var_dump would suffice. var_export just prints out valid PHP code.

|

Aswita Hidayat · Accepted Answer · 2019-09-20 07:12:04Z

2

short way, you can use Ternary Operators

$filename = !empty($_POST['filename'])?$_POST['filename']:'-';

answered Sep 20, 2019 at 7:12

Aswita Hidayat

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Comments

open source guy · Accepted Answer · 2012-12-31 05:45:40Z

1

if(!empty($_POST['filename'])){
$filename = $_POST['filename'];

echo $filename;
}

answered Dec 31, 2012 at 5:45

open source guy

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Comments

Thusitha Thilina Dayaratne · Accepted Answer · 2015-09-30 06:53:41Z

1

Please try this

error_reporting = E_ALL & ~E_NOTICE

in php.ini

edited Sep 30, 2015 at 6:53

Thusitha Thilina Dayaratne

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answered Sep 29, 2015 at 12:35

Piseth Sok

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1 Comment

Félix Adriyel Gagnon-Grenier Over a year ago

that would have been a comment

ravi404 · Accepted Answer · 2012-12-31 05:45:11Z

0

Simply

if(isset($_POST['filename'])){
 $filename = $_POST['filename'];
 echo $filename;
}
else{
 echo "POST filename is not assigned";
}

answered Dec 31, 2012 at 5:45

ravi404

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Comments

Yogesh Suthar · Accepted Answer · 2012-12-31 05:46:22Z

0

use isset for this purpose

<?php

 $index = 1;
 if(isset($_POST['filename'])) {
     $filename = $_POST['filename'];
     echo $filename;
 }

?>

answered Dec 31, 2012 at 5:46

Yogesh Suthar

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Comments

Shaunak D · Accepted Answer · 2015-05-12 13:01:13Z

0

Use empty() to check if it is available. Try with -

will generate the error if host is not present here

if(!empty($_GET["host"]))
if($_GET["host"]!="")

edited May 12, 2015 at 13:01

Shaunak D

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answered May 12, 2015 at 12:17

Danish Iqbal

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Comments

Gibolt · Accepted Answer · 2018-09-05 05:33:51Z

-1

enctype="multipart/form-data"

Check your enctype in the form before submitting

edited Sep 5, 2018 at 5:33

Gibolt

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answered Sep 5, 2018 at 4:42

Goki

1

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