Question

所以我尝试使用PHP OOP，并尝试创建一个简单的mysql数据库连接类。这是我的代码，然后我将解释发生了什么。

的index.php

<?php

// Connect to Database Class
require_once('../libs/DB.class');
$connection = new DatabaseConnect('localhost', '', '');

DB.class

<?php

class DatabaseConnect {

  // When class is called, make sure to grab the host, username, and password to connect to the MySQL Database

  public function __construct($db_host, $db_username, $db_password) {
    echo 'Attempting Connection . . . <br>';

    // If the method Connect() doesn't return true then kill the script and say you done messed up... or tell the user that connection has been successful

    if(!$this->Connect($db_host, $db_username, $db_password)) {
      die("Connection to $db_host failed");
    } else {
      echo "Connected to $db_host";
    }
  }

  // When the Connect method is called from the construct, grab the passed variables, try to connect, and return true or false

  public function Connect($db_host, $db_username, $db_password) {
    if(!mysql_connect($db_host, $db_username, $db_password)) {
      return false;
    } else {
      return true;
    }
  }

}

由于我在index.php文件中创建对象时没有提供用户名和密码，因此该构造应该杀死脚本并说我无法连接到mysql。但是，无论我向$ db_username或$ db_password变量提供什么，无论登录信息是否正确，Connect（）方法总是返回true。它永远不会返回虚假。我不知道为什么。任何帮助将不胜感激。

谢谢！

Answer 1

mysql_connect()在PHP 5.5.0中已弃用，在PHP 7.0.0中已被删除。你可以用

PDO :: __ construct（）

try {
     $dbh = new PDO($db_host, $db_username, $db_password);
    } catch (PDOException $e) {
        echo 'Connection failed: ' . $e->getMessage();
    }

<强> mysqli_connect（）

if(!mysqli_connect($db_host, $db_username, $db_password)) {
      return false;
    } else {
      return true;
    }

PHP OOP：当错误信息传递时，mysql_connect不返回false

1 个答案: