POST variable in array's name php

Question

I don't manage to get this piece of code working. I need to get an array depending of the variable posted. I guess it's obvious but I can't find the slution.

$choice1 =
    array (
        'order' => array (1,2,3,4,5),
            'settings' => (1,0,1)
    );
$choice2 =
    array (
        'order' => array (1,5,3,2,4),
            'settings' => (0,0,0)
    );
if(isset($_POST['choice'])) {
    $template_to_get = $_POST['choice'];
    $order_display = $template_to_get['order'];          // Here is the problem
    echo json_encode(array('order' => $order_display));
}

Also tried:

$order_display = $$template_to_get['order'];
$order_display = "$".$template_to_get['order'];
...

If I write this line it works but I don't know if it's choice1 or choice 2 which will be posted:

$order_display = $choice1['order'];

I would like to get the (1,2,3,4,5) array as output. (I simplified but I have around 20 choiceX)

Thanks!

smassey · Accepted Answer · 2012-06-10 17:27:01Z

2

I would group your choices up into a single num indexed array and not add the smell of variable variables:

$choices = array(
    1 => array(
        'order' => array (1,5,3,2,4),
        'settings' => (0,0,0),
    ),
    2 => array(
        'order' => array (1,2,3,4,5),
        'settings' => (1,0,1)
    ),
);

if(isset($_POST['choice'])) {
    $template_choice = $_POST['choice'];
    echo json_encode(array('order' => $choices[ $template_choice ]['order'] ));
}

Or something very close.

Edit: Note that the same would work perfectly fine if you're posting strings in $_POST['choice'], for example 'template_1' could be posted and matched in:

$choices = array(
    'template_1' => array( ... );
    ...
);

Cheers

answered Jun 10, 2012 at 17:27

smassey

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Comments

DaveRandom · Accepted Answer · 2012-06-10 14:14:26Z

1

It's a little hard to work out what you want here, but I think it's this:

$order_display = $choice1[$_POST['choice']];

answered Jun 10, 2012 at 14:14

DaveRandom

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1 Comment

remyremy Over a year ago

It's not what I want to do, I have edited my question to make it more clear

devasia2112 · Accepted Answer · 2012-06-10 14:19:41Z

0

foreach ($choice as $key => $val)
foreach ($val as $nval)
  echo $key . ' : ' . $nval . '<br />';

answered Jun 10, 2012 at 14:19

devasia2112

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Comments

Choo · Accepted Answer · 2012-06-10 14:30:05Z

0

$varname = 'choice' . $_POST['choice'];
$order_display = $$varname['order'];

edited Jun 10, 2012 at 14:30

answered Jun 10, 2012 at 14:24

Choo

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1 Comment

remyremy Over a year ago

I get: Notice: Undefined variable: c in... at the 3rd line. "c" is the first character of what I post

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