PHP - 构造函数不返回false

How can I let the $foo variable below know that foo should be false?

我怎么能让下面的$ foo变量知道foo应该是假的？

class foo extends fooBase{

  private
    $stuff;

  function __construct($something = false){
    if(is_int($something)) $this->stuff = &getStuff($something);
    else $this->stuff = $GLOBALS['something'];

    if(!$this->stuff) return false;
  }

}

$foo = new foo(435);  // 435 does not exist
if(!$foo) die(); // <-- doesn't work :(

2 个解决方案

#1

You cannot return a value from the constructor. You can use exceptions for that.

您无法从构造函数返回值。您可以使用例外。

function __construct($something = false){
    if(is_int($something)) $this->stuff = &getStuff($something);
    else $this->stuff = $GLOBALS['something'];

    if (!$this->stuff) {
        throw new Exception('Foo Not Found');
    }
}

And in your instantiation code:

在您的实例化代码中：

try {
    $foo = new foo(435);
} catch (Exception $e) {
    // handle exception
}

You can also extend exceptions.

您还可以扩展例外。

#2

Constructor is not supposed to return anything.

构造函数不应该返回任何东西。

If you need to validate data before using the to create an object, you should use a factory class.

如果在使用创建对象之前需要验证数据，则应使用工厂类。

Edit: yeah , exceptions would do the trick too, but you should not have any logic inside the constructor. It becomes a pain for unit-testing.

编辑：是的，异常也可以做到这一点，但你不应该在构造函数中有任何逻辑。它成为单元测试的痛苦。

#1