Question

我通过使用此单例类来获取此错误。请解释一下我在做错了什么？我试图通过构造函数获取sqli对象。

致命错误：从第20行的C：\ xampp \ htdocs \ docs \ users.php中的无效上下文调用私有DB :: __ construct（）

<?php

class DB
{
    #############################
    ##### Define Connection  ####
    #############################

    const HOST      = 'localhost';
    const USER      = 'root';
    const PASSWORD  = '';
    const DATABASE  = 'ali';

    #############################

    // Data Member For MySQLi Database Connection
    private static $mysqli = NULL;

    // Database Connection
    private function __construct()
    {
        @self::$mysqli = new mysqli(self::HOST, self::USER, self::PASSWORD, self::DATABASE);

        if (self::$mysqli->connect_error) {
            throw new Exception("<b>Database Connect Error: </b>" . self::$mysqli->connect_error);
        }
        if (self::$mysqli->error) {
            throw new Exception("<b>Database Select Error: </b>" . self::$mysqli->error);
        }

        echo "<span style='color: green; font-size: 18px; font-weight: bold'>Connected</span>
        <br>"; // For testing //

        return self::$mysqli;
    }

    // Class Level Static Function to Get Database Connection
    public static function get_connection()
    {
        if ( ! isset(self::$mysqli) )
        {
            self::$mysqli = new self();
        }

        return self::$mysqli;
    }
}

?>

我的第二个文件，我正在使用这个类

<?php

require_once 'DB.php';

class users extends DB
{
    public function get_results()
    {
        $sqli = DB::get_connection();

        $user_data = $sqli->query("SELECT * FROM `users`");

        while($row = $user_data->fetch_object())
        {
            echo $row->name . "<br>";
        }
    }
}

$user = new users;
try {
    $user->get_results();
} catch (Exception $e) {
    echo $e->getMessage();
}
?>

Answer 1

如果您不打算构建典型的数据库连接单例，那么下面的代码将实现您想要的目标

  class DB
  {
     const HOST = 'localhost';
     const USER = 'root';
     const PASSWORD = '';
     const DATABASE = 'yourdb';

     private static $mysqli = null;

     private function __construct() {
        //an empty private constructor to prevent creating instances of this class
     }

     public static function get_connection() {
        if(!self::$mysqli) {
           //create mysqli object here and return instead of (typically returned) self instance 
           $conn = new mysqli(self::HOST, self::USER, self::PASSWORD, self::DATABASE);

           if($conn->connect_error) {
              throw new Exception("<b>Database Connect Error: </b>" . $conn->connect_error);
           }
           if($conn->error) {
              throw new Exception("<b>Database Select Error: </b>" . $conn->error);
           }
           self::$mysqli = $conn;
        }

        return self::$mysqli;
     }
  }

除了删除extends DB之外，

用户类不需要进行任何更改，以消除错误。

更新

  class DB
  {
     const HOST = 'localhost';
     const USER = 'root';
     const PASSWORD = '';
     const DATABASE = 'yourdb';

     private $mysqli = null;

     private function __construct() {
        $conn = new mysqli(self::HOST, self::USER, self::PASSWORD, self::DATABASE);

        if($conn->connect_error) {
           throw new Exception("<b>Database Connect Error: </b>" . $conn->connect_error);
        }
        if($conn->error) {
           throw new Exception("<b>Database Select Error: </b>" . $conn->error);
        }
        $this->mysqli = $conn;
     }


     /**
      * @return DB
      */
     public static function get_connection() {
        /**
         * @static $DB DB a static variable that is not class member but will be _shared_ across all calls to  DB::get_connection()
         *
         * WTF? http://stackoverflow.com/questions/6601027/php-and-static-variables-in-object-member-functions
         */
        static $DB = null;
        if($DB == null) {
           $DB = new DB();
        }

        return $DB;
     }


     /**
      * A wrapper for internal database object/connection/resource's query method
      * @param $sql String the sql query
      *
      * @return bool|mysqli_result
      */
     public function query($sql) {
        return $this->mysqli->query($sql);
     }
  }

数据库单例错误

1 个答案: