Question

在输入数据中，我有文件名，类名和方法名。我需要在没有构造的情况下检查此类中的文件存在（完成）类存在（完成）和方法存在。有可能吗？

Answer 1

尝试使用反射：

$method = null;

try{
    $class = new ReflectionClass('Your_Class');
    $method = $class->getMethod('Your_Method');
}
catch(Exception $e){
      echo $e->getMessage();
}

if($method != null){
     // your code here;
}

Answer 2

您可以将::class（需要PHP 5.5+）附加到您的班级名称，并执行以下检查：

<?php

class Test
{
    public function someMethod()
    {
        echo 'Hello, world.';
    }
}

var_dump(method_exists(Test::class, 'someMethod'));

Answer 3

http://ideone.com/3RIOcf

class my_class_name1 {

  public function  my_method_name1() {  

  }
}

$class_name = 'my_class_name';

$method_name = 'my_method_name';

echo class_exists($class_name)? ('Class '.$class_name.' exists'):('Class '.$class_name.' does not exist');
echo "\n";
echo class_exists($class_name.'1')? ('Class '.$class_name.'1'.' exists'):('Class '.$class_name.'1'.' does not exist');
echo "\n";
echo method_exists ($class_name.'1',$method_name)? ('Method '.$method_name.' exists in class '.$class_name.'1'):('Method '.$method_name.' does not exists in class '.$class_name.'1');
echo "\n";
echo method_exists ($class_name.'1',$method_name.'1')? ('Method '.$method_name.'1'.' exists in class '.$class_name.'1'):('Method '.$method_name.'1'.'does not exists in class '.$class_name.'1');

类中存在PHP方法而不创建对象

3 个答案: