Get one value from array output?

Question

Im trying to work out how to get the value from one setting in an array.

When i do this

<?php 
$data = array($settings);
echo "<pre>";
echo print_r($data);
echo "</pre>";
?>

I get this ( but i dont want to just spit the array out like below )

Array
(
    [0] => Array
        (
            [layout] => pizza:simple
            [style] => pizza:red
        )

)
1

I tried this but no luck ( What i need is just the value of layout or style )

<?php
$data = array($settings);
$layout = $data->layout;
?>

So basically i can now use the $layout value for things i want to do on the page.

How do i do this please?

Thanks in advance :) jonny

Why do you array-ify $settings anyway? It's already an array. Access like so, $settings['layout']. — Josh
– Josh, Commented Mar 28, 2012 at 1:56

chadpeppers · Accepted Answer · 2012-03-29 00:33:10Z

1

$data[0]['layout'] or $data[0]['style']

This is an associative array so its basically an array inside of an array. So if you had another value 1 it could have a different layout and pizza and it would be accessed by $data[1]['layout'] etc

edited Mar 29, 2012 at 0:33

answered Mar 28, 2012 at 1:53

chadpeppers

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3 Comments

jonnypixel Over a year ago

Ok thats cool! thank you, it worked. Can you explain why as well. Im doing everything backwards. Learning without manuals or school. I just pick it up as i go along.

chadpeppers Over a year ago

This is an associative array so its basically an array inside of an array. So if you had another value 1 it could have a different layout and pizza and it would be accessed by $data[1]['layout'] etc

Gray Over a year ago

Please edit your answer and move your comments up into there to provide more details for posterity. Thanks.

Kishor Kundan · Accepted Answer · 2012-03-28 02:02:00Z

1

if you intend to use

$data->layout

You will have to define class and layout as its property

class Settings {
    public $layout;
    public $style;

    public function __construct($data = null) {
         $this->layout = $data['layout'];
         $this->style = $data['style'];
    }
}

// you can now use it

$data = new Settings($settings);
echo $data->layout; // this will work

answered Mar 28, 2012 at 2:02

Kishor Kundan

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