关于如何重置类的 php 4 到 5 移植问题
有人编写了以下 php4 代码，我现在正尝试将其移植到 php5：
该类是 Foo（名称已更改以保护有罪者）。 在其中一种方法中，我们将其称为 save() 类显然是这样重置的：
$this = new Foo($this->Foo_Id);
这会导致以下错误：
( ! ) Fatal error: Cannot re-assign $this in ...
我的想法是像这样修复它，但我担心它可能不一样：
$this->Foo($this->Foo_Id);
当我包含该类时，PHP 不再抛出解析/致命错误，但就像我说的，我会实现与 php4 构造相同的事情吗？
Someone wrote the following php4 code which I now am trying to port to php5:
the class is Foo (name changed to protect the guilty).
In one of the methods, we'll call it save() the class apparently is reset like this:
$this = new Foo($this->Foo_Id);
That results in the following error:
( ! ) Fatal error: Cannot re-assign $this in ...
My idea would be to fix it like this, but I am concerned that it might not be the same:
$this->Foo($this->Foo_Id);
PHP doesn't throw the parse/fatal error anymore when I include the class, but like I said, am I going to achieve the same thing as the php4 construct?
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如果不了解课程组织的背景，就很难告诉您正确的解决方案。
根据这一点，您可以执行以下操作：
调用“save”方法的代码将获取 Foo 的实例。 从外面看，这可能看起来像：
之前：
之后：
甚至：
也许这对您有帮助。
Without knowing the context in which the classes are organized its hard to tell you the right solution.
Depending on that, you could just do a:
The code which is calling the "save" method will get an instance of Foo. From the outside this could look like:
befor:
after:
or even:
maybe this helps you.
不，不会的。 这只会重新运行构造函数中的任何实际代码（顺便说一句，您可能应该将其重命名为
__construct()无论如何），而不影响实际未在那里设置的任何属性。No, it won't. That will just rerun any actual code in the constructor (which incidentally you should probably be renaming to
__construct()anyway) without affecting any properties that aren't actually set there.