How to add array to array in PHP

Question

So I thought this would be easy, but try as I might various methods of appending values to an array in PHP, I always get NULL.

$sites = array();
$sites[0] = $_POST['site0'];
foreach($sites as $site) {
    var_dump($site);
}

$_POST['site0'] is an HTML form array, containing 11 keys and values. I get a invalid argument error for line 3. Any reason why this would occur?

The way to add values to an array is like $arr[] = 'new item'; specify more dimensions depending on structure as needed e.g. $arr[0][] = .... It makes no sense at all that var_dump would print null with the above code. Is that what you really have? — Jon
– Jon, Commented Dec 11, 2011 at 16:53
are you sure there is not typo? Or missing a BIG chunk of code before the var_dump? — ajreal
– ajreal, Commented Dec 11, 2011 at 16:54
Okay, it seems that the real problem is not that the array $sites is NULL — I have fixed that, it was an error in the HTML — but rather that foreach($sites as $site) gives me "invalid argument". There are going to be more sites, and I'd like to cycle through and process them. — delwin
– delwin, Commented Dec 11, 2011 at 17:03

human · Accepted Answer · 2012-12-27 21:32:54Z

2

Probably, $_POST['site0'] is null or empty.

You could push the value to the $sites array doing so:

$sites[] = $_POST['site0'];

The value will be pushed to the end of the array.

answered Dec 27, 2012 at 21:32

human

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Comments

Nick Baluk · Accepted Answer · 2011-12-11 17:54:42Z

1

It's really work. Just check your

$_POST['site0'];

and try add

$sites[1] = 'spam';

answered Dec 11, 2011 at 17:54

Nick Baluk

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middus · Accepted Answer · 2011-12-11 17:00:09Z

0

Your code works for me (see codepad):

Code

$_POST['site0'] = 'foobar';
$sites = array();
$sites[0] = $_POST['site0'];
var_dump($sites); // returns NULL

Result

array(1) {
  [0]=>
  string(6) "foobar"
}

Try to var_dump($_POST); to see what's actually contained in $_POST.

answered Dec 11, 2011 at 17:00

middus

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Turns out the problem is slightly different than what I thought it was — I've changed the question to reflect this.