在 PHP 类中包含 MySQL 配置文件 [重复]

Question

我想在 PHP 类中包含 MySQL 配置文件。

<?php
class User {
    private $dbHost     = "localhost";
    private $dbUsername = "root";
    private $dbPassword = "";
    private $dbName     = "gic";
    private $userTbl    = 'users';

    function __construct(){
        if(!isset($this->db)){
            // Connect to the database
            $conn = new mysqli($this->dbHost, $this->dbUsername, $this->dbPassword, $this->dbName);
            if($conn->connect_error){
                die("Failed to connect with MySQL: " . $conn->connect_error);
            }else{
                $this->db = $conn;
            }
        }
    }

    function checkUser($userData = array()){
        if(!empty($userData)){

            $host  = $_SERVER['HTTP_HOST'];
            $host_upper = strtoupper($host);
            $path   = rtrim(dirname($_SERVER['PHP_SELF']), '/\\');
            $baseurl = "http://" . $host . $path ."/";


            $prevQuery = "SELECT * FROM ".$this->userTbl." WHERE oauth_provider = '".$userData['oauth_provider']."' AND oauth_uid = '".$userData['oauth_uid']."'";
            $prevResult = $this->db->query($prevQuery);

           <!-- rest of my code --> 
        }
    }
}

当我运行这个文件时，它工作正常，但我想将 MySQL 数据库连接分离到另一个文件 (config.php)，所以我编辑了我的代码：

<?php
class User {

include 'config.php';

$connection = new createCon();
$conn = $connection->connect();

function checkUser($userData = array()){
    if(!empty($userData)){

        $host  = $_SERVER['HTTP_HOST'];
        $host_upper = strtoupper($host);
        $path   = rtrim(dirname($_SERVER['PHP_SELF']), '/\\');
        $baseurl = "http://" . $host . $path ."/";


        $prevQuery = "SELECT * FROM ".$this->userTbl." WHERE oauth_provider = '".$userData['oauth_provider']."' AND oauth_uid = '".$userData['oauth_uid']."'";
        $prevResult = $conn->query($prevQuery);
        }
    }
}

和config.php

<?php

class createCon  {
    var $host = 'localhost';
    var $user = 'root';
    var $pass = '';
    var $db = 'gic';
    var $myconn;

    function connect() {
        $con = mysqli_connect($this->host, $this->user, $this->pass, $this->db);
        if (!$con) {
            die('Could not connect to database!');
        } else {
            $this->myconn = $con;
            echo 'Connection established!';}
        return $this->myconn;
    }

    function close() {
        mysqli_close($myconn);
        echo 'Connection closed!';
    }
}
?>

当我运行上面的代码时，它显示一个错误：

解析错误：语法错误、意外的“包含”(T_INCLUDE)、第 4 行 C:\xampp\htdocs\gic001\User.php 中的预期函数 (T_FUNCTION) 或 const (T_CONST)

有什么想法吗？

Answer 1

坦率地说，你的整个方法都是错误的。

A user is not a database! 用户类中不应该有 $dbHost 这样的属性。它也不应该费心连接到数据库。用户类应该只使用在别处定义的连接。

您应该创建一个数据库实例，一次。然后将它作为参数传递给用户的构造方法

class User {
    protected $conn;
    public function __construct($conn) {
        $this->conn = $conn;
    }
    function checkUser($userData = array()){
        $prevResult = $this->conn->query($prevQuery);
    }
}

所以只需包含一次您的凭据，就在创建连接的位置。

Answer 2

尝试将包含移到课程之外。

<?php

require('config.php');

class User {

否则将其放在函数内部：

class User {

 protected $conn;       


  public function __construct(){
    $this->loadConfig();
  }


  protected function loadConfig(){
     require('config.php');
     $connection = new createCon();
     $this->conn = $connection->connect();
  }

还有你想要研究的东西：

• Try / Catch 用于捕获错误的语句。
• Inheritance
• Namespaces
• Visibility

Answer 3

有了继承，应该问题不大。保持你的配置文件原样，试试这个。

include_once 'config.php';

class User extends createCon {

    private $conn;

    function __construct(){
        $this->conn = $this->connect();
    }
}