你如何在 PHP 中显式地创建一个变量的副本？
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How do you explicitly create a copy of a variable in PHP?
提问by DisgruntledGoat
I have an array of stdClass objects. When I assign one to a variable, it is not copying the variable but instead referencing the original variable. My code is like this:
我有一个 stdClass 对象数组。当我给一个变量赋值时，它不是复制变量，而是引用原始变量。我的代码是这样的：
for ( $i = 0, $len = count($rows); $i < $len; $i++ )
{
$row = $rows[$i];
echo $rows[$i]->games;
$row->games = 'test';
echo $rows[$i]->games;
}
The first echo outputs the normal value, but the second echo outputs "test". Even though I am setting the property on $row(which should be copied), it's actually setting it on the original array element.
第一个回声输出正常值，但第二个回声输出“测试”。即使我将属性设置为$row（应该复制），它实际上是在原始数组元素上设置它。
Why is this, and how do I actually create a copy, so that modifying the copy doesn't modify the original?
为什么会这样，我如何实际创建副本，以便修改副本不会修改原始副本？
回答by Pekka
Use the clonekeyword.
使用clone关键字。
$copy = clone $object;
important to note:
重要的是要注意：
When an object is cloned, PHP 5 will perform a shallow copy of all of the object's properties. Any properties that are references to other variables, will remain references.
克隆对象时，PHP 5 将执行对象所有属性的浅拷贝。任何引用其他变量的属性都将保持引用。
it comes with a nice magic method:
它带有一个很好的魔法方法：
Once the cloning is complete, if a __clone() method is defined, then the newly created object's __clone() method will be called, to allow any necessary properties that need to be changed.
克隆完成后，如果定义了 __clone() 方法，则将调用新创建的对象的 __clone() 方法，以允许更改任何必要的属性。
