Variable in an array, is this possible? syntax error probably

Question

First I declare a variable that I want to use in multiple arrays through the code. This is a snippet of it where it should work, but it doesn't:

The variable:

$test = '<div id="test"> dit is een test </div>';

The array:

$sections[] = array( 
                'title' => $test, 
                'icon' => '/img/icons/home.png' 
                );

The title is always empty :-/

Thanks !!

Can you provide more of your script? The code you are providing should work. — Richard Tuin
– Richard Tuin, Commented Jul 30, 2012 at 12:01
There is nothing wrong with your code. $sections[0]['title'] should display it fine (assuming it's the first element in the array) — amctavish
– amctavish, Commented Jul 30, 2012 at 12:01
yes thats possible, did you try doing echo $sections[0]['title']; to get title ... that should give you the title — Sudhir Bastakoti
– Sudhir Bastakoti, Commented Jul 30, 2012 at 12:01
The code is fine. Likely the array declaration is in a place that cannot read $test (i.e. a scoping issue). Either way, we'll need to see more code. — Mitya
– Mitya, Commented Jul 30, 2012 at 12:01
did you var_dump($sections) to see values stored in your array?? — Sibu
– Sibu, Commented Jul 30, 2012 at 12:10

Sumesh TG · Accepted Answer · 2012-07-30 12:12:56Z

1

Correct declaration is

$test = '<div id="test"> dit is een test </div>'; 
$sections = array( 
                'title' => $test, 
                'icon' => '/img/icons/home.png' 
                );

answered Jul 30, 2012 at 12:12

Sumesh TG

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2 Comments

user1562864 Over a year ago

Thanks for the fast response everybody. In fact, tha code is a snippet from a wordpress framework. It is one menu item in the admin panel. The above code, doesn't work, I get an error "Invalid argument supplied for foreach() in xxx/options.php" so it is more complicated than I thought.

user1562864 Over a year ago

I have found it, it was the first line of the code above that was the problem: $sections = array(); Thanks everyone !

Tiwenty · Accepted Answer · 2012-07-30 12:06:47Z

0

Try with this code :

$test = '<div id="test"> dit is een test </div>';

$sections['title'] = $test;
$sections['icon'] = '/img/icons/home.png';

echo $sections['title'];

answered Jul 30, 2012 at 12:06

Tiwenty

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