改变一个变量的值会影响另一个变量
每个人。我是 PHP 新手。我在 DateTime 方面遇到了这个问题：
$t1 = new DateTime();
$t1->setTime(9, 30);
$t2 = $t1;
$t2->add (new DateInterval('PT10M'));
echo $t1->format('H:i'); # outputs 9:40
如您所见，通过更改 $t2 的值，我还更改了 $t1 的值，这不是我想要的。请您告诉我为什么会发生这种情况，以及如何避免这种情况。谢谢。
伊恩
everyone. I am new to PHP. I am having this problem with DateTime:
$t1 = new DateTime();
$t1->setTime(9, 30);
$t2 = $t1;
$t2->add (new DateInterval('PT10M'));
echo $t1->format('H:i'); # outputs 9:40
As you can see, by changing the value of $t2, I also changed the value of $t1, which is not what I want. Would you please tell me why is this happening, and how to avoid it. Thank you.
Ian
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$t1和$t2只是对对象的引用。当您执行$t1 = $t2时，您只是复制引用，而不是对象。您可以改为
$t1 = clone $t2。$t1and$t2are just reference to an object. When you do$t1 = $t2, you are just copying the reference, not the object.You could to
$t1 = clone $t2instead.您正在使 $t2 指向 $t1。因此，编辑 $t2 会导致您编辑 $t1 的内存。
使用克隆代替：$t2 =克隆$t1
You are causing $t2 to point to $t1. So editing $t2 causes you to edit the memory of $t1.
Use clone instead: $t2 = clone $t1
你应该使用
You should use