Is it possible to define PHP variables in a for loop using $i?

Question

I am looking to do something like the following:

for ($i=0; $i<=$number_of_bundles; $i++) {

    $rug_size . $i = $_POST['Size' . $i];

}

Unfortunately this does not work since I get "undefined index" and when I try echoeing, for example, $rug_size1 is undefined.

How can something like this be done?

Whenever you think you want this, what you actually want is an array! — deceze
– deceze ♦, Commented Dec 18, 2013 at 7:29

Tobias · Accepted Answer · 2013-12-18 07:29:53Z

3

Assuming you initialize $rug_size as an array():

$rug_size[$i] = $_POST['Size' . $i];

If you really want to use different variable names (hopefully not):

$vn = 'rug_size' . $i;
$$vn = $_POST['Size' . $i];

answered Dec 18, 2013 at 7:29

Tobias

7,7711 gold badge30 silver badges44 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Raphael Rafatpanah Over a year ago

Wow, this is a very simple and smart answer. Thank you!

niyou · Accepted Answer · 2013-12-18 07:29:41Z

2

schould work like this:

for ($i=0; $i<=$number_of_bundles; $i++) {
    $varname = 'rug_size'.$i;
    $$varname = $_POST['Size'.$i];
}

http://php.net/manual/en/language.variables.variable.php

answered Dec 18, 2013 at 7:29

niyou

8791 gold badge11 silver badges23 bronze badges

Comments

Ignacio Ocampo · Accepted Answer · 2013-12-18 07:34:07Z

2

You need wrap you dynamic variable within {} (and concat string with Dot), like:

<?php
$i = 1;
${"rug_size".$i} = "Hello world!";
echo $rug_size1;
?>

Output:

Hello world!

edited Dec 18, 2013 at 7:34

answered Dec 18, 2013 at 7:28

Ignacio Ocampo

2,7231 gold badge22 silver badges31 bronze badges

2 Comments

Raphael Rafatpanah Over a year ago

That doesn't work since it is inside of a for loop. Unexpected {.

Ignacio Ocampo Over a year ago

Using ${} is a way to create dynamic variables

Chaulagai · Accepted Answer · 2013-12-18 07:47:44Z

0

do like this .

for ($i=0; $i<=$number_of_bundles; $i++) {

    ${'rug_size' . $i} = $_POST['Size' . $i];
}

answered Dec 18, 2013 at 7:47

Chaulagai

7528 silver badges12 bronze badges

Comments

R R · Accepted Answer · 2013-12-18 07:33:31Z

-1

if i am not wrong than

$_POST['Size' , $i]

will be(i.e. dot instead of comma)

$_POST['Size'. $i]

another way to avoid error:use isset()

for ($i=0; $i<=$number_of_bundles; $i++) {
if(isset($_POST['Size' . $i]))
 {
  $rug_size . $i = $_POST['Size' . $i];
 }
}

edited Dec 18, 2013 at 7:33

answered Dec 18, 2013 at 7:26

R R

2,9562 gold badges27 silver badges43 bronze badges

4 Comments

Raphael Rafatpanah Over a year ago

Sorry, for some reason I used a comma for my question. During my test I used the dot and it does not seem to work.

R R Over a year ago

@RaphaelRafatpanah are you sure u have input type's name as size1,size2 and so on?

Raphael Rafatpanah Over a year ago

Yes, I am sure. Size with a capital S.

Ken Cheung Over a year ago

$rug_size . $i (as variable) should not work.If PHP really prompts undefined index I'll suggest you to check var_dump($_POST) or wrap it with a isset() checking.

Collectives™ on Stack Overflow

Is it possible to define PHP variables in a for loop using $i?

5 Answers 5

1 Comment

Comments

2 Comments

Comments

4 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

1 Comment

Comments

2 Comments

Comments

4 Comments

Your Answer

Sign up or log in

Post as a guest

Related