PHP：如何在给定类名的情况下返回实例化的类对象？

Question

They say that eval() is evil . I want to avoid the use of the eval() line using proper PHP5 functionality. Given a class name in a static class method, how do I make it return a real object?

class Model {
  public static function loadModel($sModelPath) {
    if (!(strpos(' ' . $sModelPath, '/')>0)) {
      $sModelPath .= '/' . $sModelPath;
    }
    $sModelName = str_replace('/','_',$sModelPath);
    // P is a global var for physical path of the website
    require_once(P . '_models/' . $sModelPath . '.php');
    eval("\$oObject = new $sModelName" . '();');
    return $oObject;
  }
}

Answer 1

return new $sModelName();

You can call functions by a dynamic name as well:

$func = "foobar";
$func("baz"); //foobar("baz")

Answer 2

Yep, Kenaniah beat me to it. Gotta type faster...

More info here: http://php.net/manual/en/language.oop5.php , see the first user note .

Answer 3

I know this is a 2 year old question, but I just want to point out, you all missed the question here! All your functions do not return an instantiated class, but a new class. This includes the initial function posted from the questioner!

If you want to return an instantiated class , you have to keep track of your classes and their properties in an array, and return them from there when you require an instance of the class, instead of a re-initialised class.

This method also saves a lot of processing time if your classes do a lot of processing when they are constructed. It's always better to get an instance than to create a new class, unless you really want the class-properties re-initialized. Pay attention!

Answer 4

Try:

$m = new Model();
$m = $m->makeModel();

class Model {
  public static function loadModel($sModelPath) {
    if (!(strpos(' ' . $sModelPath, '/')>0)) {
      $sModelPath .= '/' . $sModelPath;
    }
    $sModelName = str_replace('/','_',$sModelPath);
    // P is a global var for physical path of the website
    require_once(P . '_models/' . $sModelPath . '.php');
    function makeModel(){
      $model = new $sModelName;
      return  $model;
    }
  }

}