Question

有什么方法可以使我的变量不可变，以便可以在foreach中的任何地方调用具有相同变量名的类实例吗？

我在$user->count()之后调用foreach，但$user修改了foreach变量，并引发了一些错误：

致命错误：未捕获错误：调用C：\ xampp \ htdocs \ public_html \ test.php：48中未定义的方法stdClass :: count（）堆栈跟踪：＃0 {main}抛出于C：\ xampp \ htdocs \ public_html \ test.php，第48行

实现如下：

<?php 
include 'connection.php'; 

class Collection {
  public $items;
  public function __construct ($items = []) {
    $this->items = $items;
  }
  public function get() {
    $object = $this->items;
    return $object;
  }

  public function count() {
    return count($this->items);
  }

}


class User extends Collection {

}

class Post extends Collection {

}

$stmt = $connect->prepare("SELECT * FROM users_account");
$stmt->execute();
$row = $stmt->fetchAll(PDO::FETCH_OBJ);
$user = new User($row);
$users = $user->get();

$stmt = $connect->prepare("SELECT * FROM posts");
$stmt->execute();
$row = $stmt->fetchAll(PDO::FETCH_OBJ);
$post = new Post($row);
$posts = $post->get();


foreach ($users as $user) {
  echo $user->username.'<br />';
}

echo $user->count();

echo "<hr />";

foreach ($posts as $post) {
  echo $post->body.'<br />';
}

echo $post->count();

Answer 1

为什么不简单地使用其他变量名？这是一个容易得多的解决方案，可以使代码更具可读性。

class Collection
{
    protected $items; // protected because having strong encapsulation is good

    public function __construct($items = [])
    {
        $this->items = $items;
    }

    public function get()
    {
        return $this->items; // no need to put it in a separate variable if you're simply returning
    }

    public function count()
    {
        return count($this->items);
    }
}

class User extends Collection {}

// blah blah blah

$users = new User($row);
$userList = $users->get();

foreach ($userList as $user) {
    // do work
}

使变量不可变

1 个答案: