在PHP中从一个位置建立连接

debugcn 发表于 Dev

用户名

在JSP中，如果要连接到数据库，我将使用以下代码创建一个名为DBManager的Java类：

public class DBManager {
    private final static String DB_URL = "jdbc:mysql://localhost:3306/mydb";
    private final static String DB_USERNAME = "root";
    private final static String DB_PASSWORD = "root";
    public static Connection conn = null;
    private static Statement stmt = null;

    /**
     * Tests connection with the database by getting connection using the
     * database url and username and password. And creates a dumb statement and
     * closes it to make sure everything is working fine.
     */
     static {
         try {
            Class.forName("com.mysql.jdbc.Driver");
            conn = DriverManager
                .getConnection(DB_URL, DB_USERNAME, DB_PASSWORD);
            stmt = conn.createStatement();
            stmt.close();
         } catch (Exception e) {
            e.printStackTrace();
         }
     }
}

现在，我可以轻松地执行以下操作：

PreparedStatement pstmt = DBManager.conn.prepareStatement("SELECT * FROM USER");

我自己学习PHP，而“大多数”在线教程实际上并没有教授如何以正确的方式做事。

他们都以传统方式在需要连接的每个页面中执行此操作：

$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

哪个错误，因为您说要更改密码？然后，您必须在使用该密码的每个页面中进行更改。

我的问题：我如何做与PHP中的Java类等效/相似的事情？

编辑：

<?php 
class DBManager {
   public static $conn = null;
   private static $hostname = "localhost";
   private static $username = "root";
   private static $password = "root";
   private static $dbname = "tutorme";

   protected function __construct() {
       try {
           DBManager::$conn = new PDO("mysql:host=localhost;dbname=tutorme", DBManager::$username, DBManager::$password);
           DBManager::$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
        } catch(PDOException $e) {
            echo "Error: " . $e->getMessage();
        }
    }

    public static function getInstance() {
        if (null === DBManager::$conn) {
            DBManager::$conn = new DBManager();
        }
       return DBManager::$conn;
   }
}
?>

我搜索了一下，并想出了上面的代码。但是，现在当我打电话时

$stmt = DBManager::getInstance()->prepare("INSERT INTO SUBJECT (SubjectTitle, SubjectName) VALUES (:subject,:subj)");

我收到一个错误，DBManager称没有函数prepare()

这表示我DBManager::getInstance要传回DBManager object而不是PDO conn object

423

您有一个明显的错误：

DBManager::$conn = new DBManager();

你不要那个！

只需致电new DBManager()，$connvar就会被初始化！

还有一些小的变化，

<?php 
class DBManager {
    public static $conn = null;
    private static $hostname = "localhost";
    private static $username = "root";
    private static $password = "root";
    private static $dbname = "tutorme";

    private function __construct() {
        try { // why not using $hostname and $dbname?!
           DBManager::$conn = new PDO("mysql:host=" . DBManager::$hostname . ";dbname=" .DBManager::$dbname, DBManager::$username, DBManager::$password);
           DBManager::$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
        } catch(PDOException $e) {
           echo "Error: " . $e->getMessage();
        }
    }

    // getConnection() is a better naming of the function
    public static function getConnection() {
        if (is_null(DBManager::$conn)) {
            new DBManager();
       }
       // now $conn is initialized
       return DBManager::$conn;
    }
 }
 ?>

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