Question

我知道之前曾问过类似的问题，但没有与像这样的ajax一起问过... 因此，我尝试了对数据库类进行依赖注入，并使用红色指南（例如https://phpdelusions.net/pdo/common_mistakes）声明应该建立数据库的一个连接实例并将其传递给所有类/函数...

问题是，当我使用ajax从Index.php到myfunctions.php进行发布时，该怎么办...我认为问题出在这里：

$database = new Database();
$conn = $database->getConnection();

我是否可以在Database.php中实例化它都没有关系，因为即使我有“ require_once”，每个ajax调用也会运行它...（通过简单的测试进行验证，以从Database类的构造函数中打印出日志消息）

Database.php

<?php

class Database 
{
    private $host = "";
    private $db_name = "";
    private $username = "";
    private $password = "";
    private $conn;


    function __construct()
    {
        $this->host = "localhost";
        $this->db_name = "mydb";
        $this->username = "user";
        $this->password = "pass";           
    }


    public function getConnection()
    {
        $this->conn = mysqli_connect($this->host, $this->username, $this->password, $this->db_name);

        if (mysqli_connect_errno())
        {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }

        return $this->conn;
    }   
}
?>

Index.php

<?php
header("Access-Control-Allow-Origin: *");

    // start session, and check for valid login
?>

<!DOCTYPE html>
<html>
// all HTML including bind jquery ajax calls to functions.php
$.ajax({ method: "POST", url: "myfunctions.php", }) ... and so on

MyFunctions.php

<?php
header("Access-Control-Allow-Origin: *");
header("Content-Type: application/json; charset=UTF-8");

require_once "database.php";

$database = new Database();
$conn = $database->getConnection();

MyFunctions::start();


class MyFunctions
{
    public static function start()
    {
       // Do stuff... 
       $stmt = $conn... // and so on
    }
//....

如何仅将一个数据库实例与ajax结合使用

0 个答案: