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    ArrayObject::getArrayCopy

    (PHP 5, PHP 7, PHP 8)

    ArrayObject::getArrayCopy — ArrayObject 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

    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
    8 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
    13 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
    8 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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