PHP 关闭 mysqli 与数据库的连接错误

Question

在尝试解决我的问题时，我遇到此错误的情况与我在 SO 上看到的情况不同。我有一个类，Database，它只创建它自己的一个实例，试图将与 mysql 服务器的连接数量限制为 1。这是我的类(class)[代码 1]:

class Database {

    private $_connection;

    // Store the single instance.
    private static $_instance;

    /**
     * Get self instance of database to private static variable $_instance.
     * @param string $host
     * @param string $username
     * @param string $password
     * @param string $database
     * @return Database
     */
      public static function getInstance($host,$username,$password,$database) {
          if (!self::$_instance) {
              self::$_instance = new self($host,$username,$password,$database);
          }
          return self::$_instance;
      }

      /**
       * Constructor.
       * @param string $host
       * @param string $username
       * @param string $password
       * @param string $database
       */
       public function __construct($host,$username,$password,$database) {
           $this->_connection = new mysqli($host, $username, $password, $database);

        // Error handling.
        if (mysqli_connect_error()) {
            trigger_error('Failed to connect to MySQL: ' . mysqli_connect_error(), E_USER_ERROR);
        }
    }

    /**
    * Empty clone magic method to prevent duplication.
    */
     private function __clone() {}

     /**
     * Get the mysqli connection;
     */ 
      public function getConnection(){
          return $this->_connection;
      }
  }

在这个类之后，我创建了一个连接来从表中获取一些信息。代码如下[代码2]:

    // Establish a connection with MySQL Server database.
    $host = 'localhost';
    $username = 'barbu';
    $password = 'watcrn0y';
    $database = 'siteinfo';

    $db = Database::getInstance($host, $username, $password, $database);
    $mysqli = $db->getConnection();

    // Get the firstname of the author of the site from database.
    $sql_query = 'SELECT author.firstname, author.lastname ';
    $sql_query .= 'FROM author;';

    $result = $mysqli->query($sql_query);


    if($result && $row = $result->fetch_assoc()){
        $author_firstname = $row['firstname'];
        $author_lastname = $row['lastname'];
    }

现在，在另一个文件中，我执行以下操作[代码 3]:

require '../includes/siteinfo.php'; // this file contains the connection 
    // with first database ['siteinfo']. 
    //I include this file for accessing some other variables from it, which aren't  
   //in the code posted above. 

// Establish the connection to server.
$host = 'localhost';
$username = 'barbu';
$password = 'watcrn0y';
$database = 'articles';

// Here I need to close my previous connection and begin another.
// It is important to remember that my class is designed for creating 
// just one connection at a time.
// So if I want to change the connection to another database, 
// I have to close the previous one and create the one which 
// suits my needs.
$mysqli->close();
unset($db);

$db = Database::getInstance($host, $username, $password, $database);
$mysqli = $db->getConnection();



// Send the post info to the server.
$title = (isset($_POST['title'])) ? $_POST['title'] : '';
$text = (isset($_POST['text'])) ? $_POST['text'] : '';
$sql = "INSERT INTO postInfo ";
$sql .= "VALUES ('" . $author_firstname . " " . $author_lastname ."', '" . 
date('d D M Y') . "', '" . $title . "', '" . $text . "');";
$result = $mysqli->query($sql);

执行此操作时，出现错误:

Warning: mysqli::query(): Couldn't fetch mysqli in /home/barbu/blog/admin/index.php on line 24

如果我不关闭第一个连接(假设我只让 ['unset($db)'])，我的查询将在第一个数据库 ['siteinfo'] 上执行，我得到另一条错误消息，即告诉我“siteinfo”数据库中不存在“postInfo”表，这是真的。如果我让该连接持续存在，并声明数据库类的另一个实例 $db1 和另一个 mysqli 对象 $mysqli1(它保存我的连接)，并通过它执行我的查询，我会得到与第二种情况相同的 mysqli 错误消息:“siteinfo.postInfo”不存在。你推荐我什么？我该如何解决这个问题？

Answer 1

首先，如果您希望每个 session 只有一个连接并且不允许创建第二个实例，您应该将 Database::__construct 定义为私有(private)。然后添加一个新方法Database::close。此方法的思想是关闭连接并将到类 Database 实例的链接设置为 null。代码如下所示:

public function close() 
{
   if (self::$_instance) {
       self::$_instance->getConnection()->close();
       self::$_instance = null;
   }
}

最后一点，您应该调用 $db->close(); 而不是 $mysqli->close();