Question

我还在学习OOP所以这甚至不可能（虽然我会感到惊讶），我需要一些帮助调用另一个类方法。

这是control folder中的代码，它从视图中检索详细信息以存储损坏的图书表，并需要将图书表状态更新为“已损坏”。当我运行程序时，两个表在运行时都不会更新。我需要一个很大的帮助来完成这项工作。

控制/ MarkDamagedBook.php

include('../model/Books.php');

include('../model/DamagedBook.php');
$type=$_POST['type'];
if ($type=='save')
{ 
    $DamagedBook = new DamagedBook();
    $DamagedBook->book_id=$_POST["book_id"];
    $DamagedBook->member_id=$_POST["member_id"];
    $DamagedBook->damaged_type=$_POST["cmbpro"];
    $DamagedBook->damaged_desc=$_POST["reason"];
    $DamagedBook->penalty =$_POST["penalty"];
    $DamagedBook->reason_panalty =$_POST["reason_panalty"];
    $DamagedBook->save();

    $Books = new Books(); //create an object of Books class
    $Books->book_id=$_POST["book_id"];
    $Books->updateBook();//cal a function 


     header("Location: ../view/MarkDamagedBook/new.php"); 

    }

../ model / Books.php ＆lt;书籍课

include_once "db_connection.php";
class Books extends db_connection{

public $book_id;
public $status ;

function updateBook()
{
    $status ="damaged";
    $query ="UPDATE tbl_book SET status='$status' WHERE book_id='$this->book_id'";
    mysqli_query($this->con,$query);
}

}

模型/ DamagedBook.php

 <?php
include_once "db_connection.php";
class DamagedBook extends db_connection
{
public $book_id;
public $member_id;
public $damage_type;
public $damgbook_desc;
public $status ;

function save()
{
$query= "INSERT INTO tbl_damagebook(book_id,member_id,damage_type,damgbook_desc,penalty_type,penalty_reason)VALUES('$this->book_id','$this->member_id','$this->damaged_type','$this->damaged_desc','$this->penalty','$this->reason_panalty');";
mysqli_query($this->con,$query);
}
}

db_connection.php

<?php

class db_connection {
public $con;
private $_host = "localhost";
private $_username = "root";
private $_password = "";
private $_database = "librarysys";

public function __construct()
{
$this->con = new mysqli($this->_host, $this->_username, $this->_password, $this->_database);
if(mysqli_connect_error())
    {
trigger_error("Failed to connect to MySQL: ");
    }
  }
}
?>

mvc .control类来建模，在php中调用两个类

0 个答案: