ArrayObject::getArrayCopy

(PHP 5, PHP 7, PHP 8)

ArrayObject::getArrayCopy — Creates a copy of the ArrayObject

说明

public ArrayObject::getArrayCopy(): array

Exports the ArrayObject to an array.

参数

此函数没有参数。

返回值

Returns a copy of the array. When the ArrayObject refers to an object, an array of the properties of that object will be returned.

范例

示例 #1 ArrayObject::getArrayCopy() example


<?php
// Array of available fruits
$fruits = array("lemons" => 1, "oranges" => 4, "bananas" => 5, "apples" => 10);

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

// create a copy of the array
$copy = $fruitsArrayObject->getArrayCopy();
print_r($copy);

?>

以上例程会输出：

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

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User Contributed Notes 5 notes

down

Ivo von Putzer ¶

10 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!

down

jlshor at buffalo dot edu ¶

5 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?

down

sorcerer ¶

5 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?

down

spidgorny at gmail dot com ¶

5 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.

down

php at webflips dot net ¶

8 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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官方地址：https://www.php.net/manual/en/arrayobject.getarraycopy.php

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ArrayObject::getArrayCopy

说明

参数

返回值

范例

User Contributed Notes 5 notes