Question

这里我想将$ user_id从editdata（）函数传递给statusid（）函数，但两者都在同一个名为User的类下。

class User{

    public function __construct()
    {
        $this->db=mysqli_connect("localhost","root","","hep1");

    }
    public function execute_query($db,$query)
    {
        return mysqli_query($db,$query);

    }
public function editdata($user_id)
    {

        $query="SELECT * FROM employee_form WHERE id='$user_id'";
        $kal=$this->execute_query($this->db,$query);

        return $kal;
    }
public function statusid()
    {
        self::editdata($user_id);    //line 108

        $user_id=$user_id;   //line 110

        $query="SELECT status_name FROM employee_form WHERE id='$user_id'";
        $kal=$this->execute_query($this->db,$query);
    }
}

但是无论何时运行上面的代码，它都会显示错误，如

注意：未定义的变量：user_id in   第108行的C：\ xampp \ htdocs \ HT \ Tasks \ to-do-using-class \ class.php

注意：未定义的变量：user_id in   第110行的C：\ xampp \ htdocs \ HT \ Tasks \ to-do-using-class \ class.php

请帮助解决这个问题

Answer 1

您可以按照以下方式使用getter和setter方法。 $ user_id由表单数据设置，或者您可以作为静态传递。

<?php
    class User
    {
        protected $user_id;
        public function __construct()
        {
            $this->db = mysqli_connect("localhost", "root", "", "hep1");

        }
        public function setUserID($user_id)
        {
            return $this->user_id = $user_id;
        }
        public function getUserID()
        {
            return $this->user_id;
        }

        public function execute_query($db, $query)
        {
            return mysqli_query($db, $query);

        }
        public function editdata($user_id)
        {

            $query = "SELECT * FROM employee_form WHERE id='$user_id'";
            $kal   = $this->execute_query($this->db, $query);
            return $kal;
        }
        public function statusid()
        {
            $user_id =  self::getUserID();
            self::editdata($user_id); //line 108
            $query   = "SELECT status_name FROM employee_form WHERE id='$user_id'";
            $kal     = $this->execute_query($this->db, $query);
        }
    }

    // create a object 
    $user =  new User();
    // set user id here when you get
    $user_id = '1';
    $user->setUserID($user_id);
    $user->statusid();

Answer 2

如果要编写更多或更少的面向对象代码，则应使用面向对象的原则。现在，您的代码看起来像过程代码，但包含在类中。您可以从以下结构开始：

class UserStorage
{
    private $db;

    // Do not create connection, but instead inject it
    public function __construct($db)
    {
        $this->db = $db;
    }

    // Use interanl $db, do not pass new connection.
    public function query($query)
    {
        return mysqli_query($this->db, $query);
    }

    public function fetch($userId)
    {
        // Consider using prepared statements here.
        // Also consider using PDO and not mix procedural and OO style.
        return mysqli_fetch_assoc($this->query("SELECT * FROM employee_form WHERE id='$userId'"));
    }
}

$db = mysqli_connect("localhost", "root", "", "hep1");
$userStorage = new UserStorage($db);
$userId = 1;
$user = $userStorage->fetch($userId);
$status = $user['status_name'] ?? null;

首先阅读SOLID principles。

如何将参数从一个函数传递到另一个函数但在php中的同一个类中

2 个答案: