PHP 2d Array Declaration

Question

I am trying to define a 2d array in php. I have some concept code so you can see the situation:

    class Testing { 
        protected $requiredFieldsByReferenceType = array(
           ['Book']['volume'] => true,
           ['Book']['source'] => true,
           ['Book Section']['volume'] => true,
           ['Book Section']['source'] => true,
           ['Chart or Table']['volume'] => true,
           ['Chart or Table']['source'] => true
        );
        print_r($requiredFieldsByReferenceType);
     }//End Testing

The error that is thrown:

Parse error: syntax error, unexpected '[', expecting ')'

I kept reading the title as "deceleration". I think the misplaced E (it's "declAration") threw me off. — Frank Farmer
– Frank Farmer, Commented Jun 5, 2009 at 2:42

Itay Moav -Malimovka · Accepted Answer · 2009-06-05 02:35:45Z

5

The other answers are good.
The syntax using array() is:

$requiredFieldsByReferenceType = array('Book'=>array('volume' => true,
                                                     'source' => true),
                                       'Book Section'=>array('volume' => true,
                                                             'source' => true)
                                       );

answered Jun 5, 2009 at 2:35

Itay Moav -Malimovka

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Comments

Gumbo · Accepted Answer · 2009-06-06 16:00:23Z

5

You have to use array() inside the array value declarations too:

protected $myArray = array(
    "Book" => array(
        "item1" => true,
        "item2" => true
    ),
    "Chest" => array(
        "item1" => true,
        "item2" => false
    )
);

answered Jun 6, 2009 at 16:00

Gumbo

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2 Comments

Diego Over a year ago

How would I access it then? The normal way? $myArray['Book']['item1'] = false?

Gumbo Over a year ago

@Diego: Yes. After you’ve build that array, you can access it via the $variable[…] syntax.

Alex Weinstein · Accepted Answer · 2009-06-05 02:28:08Z

1

$requiredFieldsByReferenceType ['Book']['volume'] = true;
$requiredFieldsByReferenceType ['Book']['source'] = true;
$requiredFieldsByReferenceType ['Book Section']['volume'] = true;

answered Jun 5, 2009 at 2:28

Alex Weinstein

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Comments

grantwparks · Accepted Answer · 2009-06-05 03:40:25Z

1

Only the answers that assign the array in ONE statement are going to work in your context (defining a class property) unless you put them all in the constructor. By the same token, I don't think that print_r is going to work without being in a method...

answered Jun 5, 2009 at 3:40

grantwparks

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Comments

Yogu · Accepted Answer · 2012-07-23 18:18:19Z

1

$arr1 = array(
    array(value1,value2,value3),
    array(value4,value5,value6),
    array(value7,value8,value9),
);

edited Jul 23, 2012 at 18:18

Yogu

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answered Apr 20, 2012 at 9:37

shrwan kumar

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Comments

skoob · Accepted Answer · 2009-06-05 02:36:42Z

0

Another way to do it is by nesting array() functions:

 $requiredFieldsByReferenceType = array(
           'Book' => array('volume' => true,
                                       'source' => true),
           'Book Section' => array('volume' => true,
                                                     'source' => true),
         ...
        );

answered Jun 5, 2009 at 2:36

skoob

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Comments

cletus · Accepted Answer · 2009-06-06 16:01:53Z

0

PHP doesn't do multi-dimensional arrays. You must construct it as arrays of arrays.

protected $myArray = array(
  'Book' => array(
    'item1' => true,
    'item2' => true,
  ),
  'Chest' => array(
  ),
    'item1' => true,
    'item2' => false,
);

answered Jun 6, 2009 at 16:01

cletus

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Comments

Oli · Accepted Answer · 2009-06-06 16:02:50Z

0

Class TestClass {
    protected $myArray = array(
        "Book" => array('item1' => true, 'item2' => false),
        "chest" => array('item1' => true, 'item2' => false),
    );
}

answered Jun 6, 2009 at 16:02

Oli

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Comments

Patrick Glandien · Accepted Answer · 2009-06-06 16:04:08Z

0

The syntax to define an array is array(key0 => value0, key1 => value1, key2 => value2, ...). Since a two-dimensional array in PHP is just multiple arrays as values in an array, it would look like this:

$myArray =
    array(
        'Book' =>
            array(
                'item1' => true,
                'item2' => true
            ),
        'Chest' =>
            array(
                'item1' => true,
                'item2' => false
            )
    );

answered Jun 6, 2009 at 16:04

Patrick Glandien

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Comments

Boris Guéry · Accepted Answer · 2009-06-06 16:04:14Z

Do this instead:

    $myArray = array(
            'book' => array(
                'item1' => true,
                'item2' => true
            ),
            'chest' => array(
                'item1' => true,
                'item2' => true,
            )
        );

By the way you shouldn't initialize your attribues like this. Rather use getters/setters.

class TestClass {
    protected $_myArray;

    public function __construct()
    {
        $this->setMyArray();
    }

    public function setMyArray()
    {
        $this->_myArray = array(
            'book' => array(
                'item1' => true,
                'item2' => true
            ),
            'chest' => array(
                'item1' => true,
                'item2' => true,
            )
        );
    }
}


$foo = new TestClass();
print_r($foo);

Greg B · Accepted Answer · 2009-06-06 16:04:21Z

0

It looks like you're mixing styles of appending to an array here.

try

$arr = array(
    'key' => array ('key2' => 'value 1', 'key3' => 'value2'),
    'key12' => array('key4' => 'value4')
);

or

$arr = array();

$arr['key1'] = array();
$arr['key2'] = array();

$arr['key1']['key3'] = 'value1';

(Please note my examples don't produce the same data structure, I was just demonstrating the different methods)

answered Jun 6, 2009 at 16:04

Greg B

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