用于检查

Question

如何将类型传递给函数，以便检查对象是否来自传递的类型。

我的尝试似乎无法奏效：

class Test {
    function __construct() {

    }
}
function check(Type) {
    $x = new Test();
    if ($x instanceof Type) {
        print_r("the object is from the type of the passed variable! :)");
    }
}
check(Test);

Answer 1

如果要像上面那样实例化类型，可以将类名作为字符串传递。

function check($class_str)
{
    $x = new $class_str();
    if($x instanceof Test) {
        print_r("the object is from the type of the passed variable! :)");
    }
}
check('Test');

Answer 2

我不确定我是否正确地理解了您的问题，但如果我说得对，我就是这样做的列表对象，而不是只存储一个特定的类型或对象：

class DList extends DVector {
/**
 * @param string $template The name of the type to store
 */
public function __construct($template){
    parent::__construct();
    if(!class_exists($template)){
        if(in_array($template, array(self::Bool, self::Integer, self::String, self::Float, self::Map))){
            $this->_template = $template;
            $this->_callback = '_isTemplateType';
        } else {
            throw new DListTemplateTypeException('Unable to find the class "' . $template. '" ');
        }
    } else {
        $this->_template = $template;
        $this->_callback = '_isTemplateInstance';
    }
}

private function _isTemplateType($value){
    return gettype($value) == $this->_template;
}

private function _isTemplateInstance($value){
    return $value instanceof $this->_template;
}
}

Answer 3

function check($type) {
    $x = new Test();
    if (is_subclass_of($x, $type)) {
        print_r("the object is from the type of the passed variable! :)");
    }
}
check("Test");

http://www.php.net/manual/en/function.is-subclass-of.php