Question

如何使用__cal生成函数名称？

示例：

public function setTitleEn($title){
  $this->titleEn = $title;
}
public function setTitleDe($title){
  $this->titleDe = $title;
}
public function setBodyEn($body){
      $this->bodyEn = $body;
    }
public function setBodyDe($body){
  $this->bodyDe = $body;
}

public function __call($name, $arguments)
{
  //execution function
}

$article = new Article();
$article->setTitle("My title", "en") //execution setTitleEn
$article->setTitle("My title", "de") //execution setTitleDe
$article->setBody("My title", "de") //execution setBodyEn

等

Answer 1

使用call_user_func_array。你必须在这里建立一些假设，比如最终的论证总是被视为语言。

public function __call($name, $arguments) {
    $lang = 'en';
    if (count($arguments) > 1)
        $lang = array_shift($arguments);

    $fname = $name.ucwords($lang);
    if (method_exists($this, $fname) === false) {
        throw new Exception('Method "'.$name.'" does not exist.');
    }

    return call_user_func_array(
        array($this, $fname), 
        $arguments
    );
}

<强>文档

call_user_func_array - http://php.net/manual/en/function.call-user-func-array.php
method_exists - http://php.net/manual/en/function.method-exists.php

Answer 2

它不一定＆＃34;生成＆＃34;方法，但您可以使用__call（）

动态调用这些方法

public function __call($name, $arguments)
{
    $method = $name . ucfirst($arguments[1]);
    return $this->$method($arguments[0]);
}

PHP __call并生成函数名称

2 个答案: