Question

示例：

Class Test {
    private function __construct() {}

    public static function init() {
        $new_test = new Test();
        return $new_test->inner_test;
    }

    public function inner_test() {
        print '!!!!';
    }
}

$test = Test::init();
$test();

返回“PHP致命错误：函数名称必须是字符串”

有没有办法在PHP中完成这种Javascript类型行为？

Answer 1

PHP 5.4以下

Class Test {
    private function __construct() {}

    public static function init() {
        $new_test = new Test();
        return $new_test->inner_test();
    }

    public function inner_test() {
        print '!!!!';
    }
}

$test = array('Test','init');
call_user_func($test);

菲尔：

function test(){
echo "Hello";
}
$test = test;
$test();

在沙盒中运行它。

Answer 2

感谢Phil的链接。这有效：

Class Test {
    private function __construct($myString) {
        $this->myString = $myString;
    }

    public static function init($myString) {
        $new_test = new Test($myString);
        return function() use (&$new_test) {$new_test->inner_test();};
    }

    public function inner_test() {
        print $this->myString;
    }
}

$test = Test::init('Lorem Ipsum Dolar');
$test();

返回对方法的引用？

2 个答案: