从__construct函数声明对象的属性(declare properties of an object out of __construct function)

 我想知道声明一个对象的属性是否是一个好习惯，如： 
 
$this->name = $name;
 
 超出函数__construct 
 
 我正在尝试使用数据库表中的数据构建一个对象。 但是只有在注册了id时才会构建此对象。 我知道__construct函数总是返回一个对象，所以我不能得到错误的返回。 所以我尝试了以下内容： 
 
//test.php

$mod = new item($id);
if($mod->validate()) {
$item = $mod;
}

class item {

  protected $id;
    public function __construct($id)    {
        $this->id = $id;
    }

public function validate() {

        $db = new db('restaurants_items_modifiers');

        if($mod = $db->get($this->id)) {
            $this->price = $mod['price'];
            $this->name = $mod['name'];
            $this->desc = $mod['desc'];
            return true;
        } else {
            return false;
        }

    }
}
 
 这样可行，但这样做是一个很好的做法吗？ 或者我应该在__construct函数上声明一切？ 

I would like to know if is a good practice to declare a property of an object like:
 
$this->name = $name;
 
out of the function __construct
 
I am trying to build an object with data from a database table. But this object will be build only if the id is registered. I know that the __construct function always return an object so I can not get a false return. So I tried the following:
 
//test.php

$mod = new item($id);
if($mod->validate()) {
$item = $mod;
}

class item {

  protected $id;
    public function __construct($id)    {
        $this->id = $id;
    }

public function validate() {

        $db = new db('restaurants_items_modifiers');

        if($mod = $db->get($this->id)) {
            $this->price = $mod['price'];
            $this->name = $mod['name'];
            $this->desc = $mod['desc'];
            return true;
        } else {
            return false;
        }

    }
}
 
This will work but is a good practice to do it like this? or I should declare everything on the __construct function?

原文：https://stackoverflow.com/questions/44534813