ArrayObject::getArrayCopy

(PHP 5, PHP 7, PHP 8)

ArrayObject::getArrayCopyArrayObject のコピーを作成する

説明

public ArrayObject::getArrayCopy(): array

ArrayObject を配列にエクスポートします。

パラメータ

この関数にはパラメータはありません。

戻り値

配列のコピーを返します。ArrayObject がオブジェクトを参照している場合は、 オブジェクトのプロパティの配列を返します。

例1 ArrayObject::getArrayCopy() の例

<?php
// フルーツの配列
$fruits = array("lemons" => 1, "oranges" => 4, "bananas" => 5, "apples" => 10);

$fruitsArrayObject = new ArrayObject($fruits);
$fruitsArrayObject['pears'] = 4;

// 配列のコピーを作成します
$copy = $fruitsArrayObject->getArrayCopy();
print_r($copy);

?>

上の例の出力は以下となります。

Array
(
    [lemons] => 1
    [oranges] => 4
    [bananas] => 5
    [apples] => 10
    [pears] => 4
)

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User Contributed Notes 5 notes

up
3
jlshor at buffalo dot edu
7 years ago
Is there a difference between casting to an array and using this function?

For instance, if we have:
$arrayObject = new ArrayObject([1, 2, 3]);

Is there a difference between these:
$array = (array) $arrayObject;
vs
$array = $arrayObject->getArrayCopy();

If not, is there any scenario where they would produce different results, or do they produce the result in different ways?
up
3
Ivo von Putzer
12 years ago
If you did something like this to make your constructor multidimensional capable you will have some trouble using getArrayCopy to get a plain array straight out of the method:
<?php
public function __construct( $array = array(), $flags = 2 )
{
   
// let’s give the objects the right and not the inherited name
   
$class = get_class($this);

    foreach(
$array as $offset => $value)
       
$this->offsetSet($offset, is_array($value) ? new $class($value) : $value);

   
$this->setFlags($flags);
}
?>

That’s the way I solved it:

<?php
public function getArray($recursion = false)
{
   
// just in case the object might be multidimensional
   
if ( $this === true)
        return
$this->getArrayCopy();

    return
array_map( function($item){
        return
is_object($item) ? $item->getArray(true) : $item;
    },
$this->getArrayCopy() );
}
?>

Hope this was useful!
up
1
spidgorny at gmail dot com
7 years ago
<?php
$data
= $likeArray->getArrayCopy();
?>
will NOT be magically called if you cast to array. Although I've expected it.
<?php
$nothing
= (array)$likeArray;
?>
Here, $data != $nothing.
up
0
sorcerer
7 years ago
When I used print_r ($fruitsArrayObject) instead of print_r ($copy), i.e. ignoring the getArrayCopy() step, I still got the same output. Why?
up
0
php at webflips dot net
10 years ago
"When the ArrayObject refers to an object an array of the public properties of that object will be returned."

This description does not seem to be right:

<?php
class A
{
    public
$var = 'var';
    protected
$foo = 'foo';
    private
$bar = 'bar';
}

$o = new ArrayObject(new A());
var_dump($o->getArrayCopy());

/*
Dumps:

array(3) {
  ["var"]=>
  string(3) "var"
  ["*foo"]=>
  string(3) "foo"
  ["Abar"]=>
  string(3) "bar"
}
*/
?>

So it does not only include the public properties.
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