PHP array as variable?

Question

I'm new to PHP, I have this code:

if(!$form['profile_hunter']['field_profile_hunter_location']['und']['0']['value']['#default_value']){
    $form['profile_hunter']['field_profile_hunter_location']['und']['0']['value']['#default_value'] = $default_city;
  }

it's working but I don't like it to be that long, so I change:

$form_location = $form['profile_hunter']['field_profile_hunter_location']['und']['0']['value']['#default_value'];
if(!$form_location){
  $form_location = $city;
}

Then it's not working, why?

Ellaborate on "not working"...

prodigitalson
– prodigitalson

2012-12-16 04:34:58 +00:00
Commented Dec 16, 2012 at 4:34 — prodigitalson
– prodigitalson, Commented Dec 16, 2012 at 4:34

Supericy · Accepted Answer · 2012-12-16 05:19:44Z

3

It's because when you assign $form_location, it is making a copy of the data. In order for both variables to "point" to the same data, you would need to use the reference operator, example:

$var = &$some_var;

and in your case:

$form_location = &$form['profile_hunter']['field_profile_hunter_location']['und']['0']['value']['#default_value'];
if(!$form_location){
  $form_location = $default_city;
}

http://php.net/manual/en/language.references.php

edited Dec 16, 2012 at 5:19

answered Dec 16, 2012 at 4:36

Supericy

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Comments

Antimony · Accepted Answer · 2012-12-16 04:35:24Z

1

Because your code is assigning to $form_location, but not the actual value in the array.

The assignment makes $form_location refer to something different. The fact that its former value happened to be copied out of an array is irrelevant.

In C/C++, you could do something like this using pointers, but most higher level languages don't support it since it tends to be error prone.

Anyway, you could set a variable to the innermost array, since arrays are stored by reference. This would reduce the amount of code you need, while avoiding the problems introduced by taking a reference directly to an array element.

answered Dec 16, 2012 at 4:35

Antimony

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donutdan4114 · Accepted Answer · 2012-12-16 04:38:44Z

0

$form_location = $form['profile_hunter']['field_profile_hunter_location']['und'][0]['value']['#default_value'];
if(empty($form_location)){
  $form['profile_hunter']['field_profile_hunter_location']['und'][0]['value']['#default_value'] = $city;
}

You should probably use 'empty', that is a Drupal convention. Also "0" is not a string, but a number, so you don't need quotes around it.

answered Dec 16, 2012 at 4:38

donutdan4114

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Hana J. Chang · Accepted Answer · 2012-12-16 04:52:16Z

0

Got the answer! Thanks to Tony!

It should be

$form_location = &$form['profile_hunter']['field_profile_hunter_location']['und']['0']['value']['#default_value'];

The "&" means to pass by reference, without it would be to pass by value.

answered Dec 16, 2012 at 4:52

Hana J. Chang

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