通过值的PHP复制数组元素，而不是引用 [英] php copying array elements by value, not by reference

问题描述

我有以下的code：

$data['x'] = $this->x->getResults();  

$data['y'] = $data['x'];

//some code here to modify $data['y']
//this causes (undesirably) $data['x] to be modified as well

我猜自$数据中的所有元素本身的引用，修改$数据['Y']还修改$数据['X'] ..这不是我想要的。我想$数据['X']保持不变。有什么办法取消引用这里的要素，这样我可以通过价值元素复制？

I guess since all the elements of $data are themselves references, modifying $data['y'] also modifies $data['x']..which is NOT what I want. I want $data['x'] to remain the same. Is there any way to dereference the elements here so that I can copy the elements by value?

感谢。

更新函数：$ this-> X-> getResults（）;返回一个对象数组。所以，我就可以这样做：$数据['X'] [0] - > date_create ...

Update: $this->x->getResults(); returns an object array. So I can then do something like: $data['x'][0]->date_create ...

更新：
克隆阵列我的最新尝试，看起来是这样的：

Update: my latest attempt to clone the array looks something like this:

   $data['x'] = $this->x->getResults();     
   $data['y'] = $data['y'];
   foreach($data['x'] as $key=>$row) {
       $data['y'][$key]->some_attr = clone $row->some_attr;
   }

我是在这里下车的方式？我不断收到错误呼吁非对象__clone方法。从阅读的反应好像我最好的选择是遍历每个元素并克隆它（这是我试图与code ++做的..）。

Am I way off here? I keep getting a "__clone method called on non-object" error. From reading the responses it seems like my best option is to iterate over each element and clone it (which is what I was trying to do with that code..).

更新：只要解决了它！：foreach循环里面我只需要将该行更改为：

UPDATE: Just solved it!: inside the foreach loop I just needed to change the line to:

$data['y'][$key] = clone $row;

和它的作品！感谢大家的帮助。

And it works! Thanks to everyone for the help.

推荐答案

您可以采取的事实，即PHP将取消引用函数调用的结果。

You can take advantage of the fact that PHP will dereference the results of a function call.

下面是一些例子code我掀起了：

Here's some example code I whipped up:

$x = 'x';
$y = 'y';
$arr = array(&$x,&$y);
print_r($arr);

echo "<br/>";
$arr2 = $arr;
$arr2[0] = 'zzz';
print_r($arr);
print_r($arr2);

echo "<br/>";
$arr2 = array_flip(array_flip($arr));
$arr2[0] = '123';
print_r($arr);
print_r($arr2);

结果是这样的：

Array ( [0] => x [1] => y )
Array ( [0] => zzz [1] => y ) Array ( [0] => zzz [1] => y )
Array ( [0] => zzz [1] => y ) Array ( [0] => 123 [1] => y ) 

您可以看到，使用的结果 array_flip（） $改编的assigment期间 $ ARR2 在随后的变化差异的结果 $ ARR2 ，因为 array_flip（）要求强制取消引用。

You can see that the results of using array_flip() during the assigment of $arr to $arr2 results in differences in the subsequent changes to $arr2, as the array_flip() calls forces a dereference.

这似乎不是非常有效的，但如果 $这个 - ＆GT它可能为你工作; X-＆GT; getResults（）将返回数组：

It doesn't seem terribly efficient, but it might work for you if $this->x->getResults() is returning an array:

$data['x'] = array_flip(array_flip($this->x->getResults()));
$data['y'] = $data['x'];

请参阅本（未）线程另一个例子。

如果一切您返回数组中的一个对象然而，然后复制对象的唯一方法是使用的clone（），你将不得不通过<迭代code> $数据['X'] ，每个元素克隆到 $数据['Y'] 。

If everything in your returned array is an object however, then the only way to copy an object is to use clone(), and you would have to iterate through $data['x'] and clone each element into $data['y'].

例如：

$data['x'] = $this->x->getResults();
$data['y'] = array();
foreach($data['x'] as $key => $obj) {
    $data['y'][$key] = clone $obj;
}

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