你不能在一个PHP类中放入两个具有唯一参数签名的__construct函数。我想这样做:

class Student 
{
   protected $id;
   protected $name;
   // etc.

   public function __construct($id){
       $this->id = $id;
      // other members are still uninitialized
   }

   public function __construct($row_from_database){
       $this->id = $row_from_database->id;
       $this->name = $row_from_database->name;
       // etc.
   }
}

PHP中最好的方法是什么?


当前回答

正如这里已经展示的,在PHP中有许多方法来声明多个构造函数,但是没有一种方法是正确的(因为PHP技术上不允许这样做)。 但这并不能阻止我们破解这个功能…… 下面是另一个例子:

<?php

class myClass {
    public function __construct() {
        $get_arguments       = func_get_args();
        $number_of_arguments = func_num_args();

        if (method_exists($this, $method_name = '__construct'.$number_of_arguments)) {
            call_user_func_array(array($this, $method_name), $get_arguments);
        }
    }

    public function __construct1($argument1) {
        echo 'constructor with 1 parameter ' . $argument1 . "\n";
    }

    public function __construct2($argument1, $argument2) {
        echo 'constructor with 2 parameter ' . $argument1 . ' ' . $argument2 . "\n";
    }

    public function __construct3($argument1, $argument2, $argument3) {
        echo 'constructor with 3 parameter ' . $argument1 . ' ' . $argument2 . ' ' . $argument3 . "\n";
    }
}

$object1 = new myClass('BUET');
$object2 = new myClass('BUET', 'is');
$object3 = new myClass('BUET', 'is', 'Best.');

使用和理解多个构造函数的最简单方法:

希望这能有所帮助。:)

其他回答

你可以这样做:

public function __construct($param)
{
    if(is_int($param)) {
         $this->id = $param;
    } elseif(is_object($param)) {
     // do something else
    }
 }
public function __construct() {
    $parameters = func_get_args();
    ...
}

$o = new MyClass('One', 'Two', 3);

现在$ parameters将是一个值为'One', 'Two', 3的数组。

编辑,

我可以加上

func_num_args()

会给出函数的参数个数。

我知道我在这方面非常晚,但我提出了一个相当灵活的模式,应该允许一些真正有趣和通用的实现。

像往常一样设置类,使用您喜欢的任何变量。

class MyClass{ protected $myVar1; protected $myVar2; public function __construct($obj = null){ if($obj){ foreach (((object)$obj) as $key => $value) { if(isset($value) && in_array($key, array_keys(get_object_vars($this)))){ $this->$key = $value; } } } } } When you make your object just pass an associative array with the keys of the array the same as the names of your vars, like so... $sample_variable = new MyClass([ 'myVar2'=>123, 'i_dont_want_this_one'=> 'This won\'t make it into the class' ]); print_r($sample_variable); The print_r($sample_variable); after this instantiation yields the following: MyClass Object ( [myVar1:protected] => [myVar2:protected] => 123 ) Because we've initialize $group to null in our __construct(...), it is also valid to pass nothing whatsoever into the constructor as well, like so... $sample_variable = new MyClass(); print_r($sample_variable); Now the output is exactly as expected: MyClass Object ( [myVar1:protected] => [myVar2:protected] => ) The reason I wrote this was so that I could directly pass the output of json_decode(...) to my constructor, and not worry about it too much. This was executed in PHP 7.1. Enjoy!

正如这里已经展示的,在PHP中有许多方法来声明多个构造函数,但是没有一种方法是正确的(因为PHP技术上不允许这样做)。 但这并不能阻止我们破解这个功能…… 下面是另一个例子:

<?php

class myClass {
    public function __construct() {
        $get_arguments       = func_get_args();
        $number_of_arguments = func_num_args();

        if (method_exists($this, $method_name = '__construct'.$number_of_arguments)) {
            call_user_func_array(array($this, $method_name), $get_arguments);
        }
    }

    public function __construct1($argument1) {
        echo 'constructor with 1 parameter ' . $argument1 . "\n";
    }

    public function __construct2($argument1, $argument2) {
        echo 'constructor with 2 parameter ' . $argument1 . ' ' . $argument2 . "\n";
    }

    public function __construct3($argument1, $argument2, $argument3) {
        echo 'constructor with 3 parameter ' . $argument1 . ' ' . $argument2 . ' ' . $argument3 . "\n";
    }
}

$object1 = new myClass('BUET');
$object2 = new myClass('BUET', 'is');
$object3 = new myClass('BUET', 'is', 'Best.');

使用和理解多个构造函数的最简单方法:

希望这能有所帮助。:)

PHP是一种动态语言,因此不能重载方法。你必须像这样检查参数的类型:

class Student 
{
   protected $id;
   protected $name;
   // etc.

   public function __construct($idOrRow){
    if(is_int($idOrRow))
    {
        $this->id = $idOrRow;
        // other members are still uninitialized
    }
    else if(is_array($idOrRow))
    {
       $this->id = $idOrRow->id;
       $this->name = $idOrRow->name;
       // etc.  
    }
}