OOP mysqli的数据库功能

Question

我在PHP OOP初学者，我想我的代码玩。所以我想要做的第一件事是写一个基本的类与连接和插入数据库的功能。所以我期望的情况是这样的：

-I想上一流的，这将CONTROLL连接并插入函数创建。问题是，我的$ connect变量不能用另一个函数工作，所以我能做些什么来实现这一点？ 您将通过提供的代码了解更多。

<?php 

class DB { 
protected $dbhost = 'localhost'; 
protected $dbuser = 'root'; 
protected $dbpass = ''; 
protected $dbname = 'newdb'; 

public function connect() 
{ 
    $connect = new mysqli($this->dbhost, $this->dbuser, $this->dbpass, $this->dbname); 

    if($connect->error) 
    { 
     echo "Failed to connect"; 
    } 
    else 
    { 
     echo "connected"; 
    } 
} 

public function insert($name, $second) 
{ 
    $insert = "INSERT INTO posts (name, second) VALUES ('$name', '$second')"; 

    if ($connect->query($insert) === TRUE) 
    { 
     echo "New record created successfully"; 
    } 
    else 
    { 
     echo "Error: " . $sql . "<br>" . mysqli_connect_error(); 
    } 
} 
} 
require_once 'classes/DB.php'; 

$db = new DB; 

if(isset($_POST['submit'])) 
{ 
    $name = $_POST['name']; 
    $second = $_POST['second']; 

    $db->insert($name, $second); 
} 

?> 


<form method="POST"> 
    Add smth<input type="text" name="name"><br> 
    Also omg<input type="text" name="second"><br> 
    <input type="submit" name="submit"> 
</form> 

Answer 1

添加$connect为您的类的属性，所以你可以在类中使用$this->connect到处重用：

class DB 
{ 
    protected $dbhost = 'localhost'; 
    protected $dbuser = 'root'; 
    protected $dbpass = ''; 
    protected $dbname = 'newdb'; 
    protected $connect; 

    public function connect() 
    { 
     $this->connect = new mysqli($this->dbhost, $this->dbuser, $this->dbpass, $this->dbname); 

     if ($this->connect->error) 
     { 
      echo "Failed to connect"; 
     } 
     else 
     { 
      echo "connected"; 
     } 
    } 

    public function insert($name, $second) 
    { 
     $insert = "INSERT INTO posts (name, second) VALUES ('$name', '$second')"; 

     if ($this->connect->query($insert) === TRUE) 
     { 
      echo "New record created successfully"; 
     } 
     else 
     { 
      echo "Error: " . $sql . "<br>" . mysqli_connect_error(); 
     } 
    } 
} 

Answer 2

我已经重写你的类了一下，这样你就可以将你的数据库在构造函数中的设置。所以它确实是一个对象，你可以使用多个数据库。在你的类中你做错了什么是使用变量$ connect，而不是将它声明为整个类中可用的变量。

类

<?php 

class DB { 
    protected $dbhost; 
    protected $dbuser; 
    protected $dbpass; 
    protected $dbname; 
    protected $connection; 

    public function __construct($dbhost, $dbuser, $dbpass, $dbname) 
    { 
     $this->dbhost = $dbhost; 
     $this->dbuser = $dbuser; 
     $this->dbpass = $dbpass; 
     $this->dbname = $dbname; 

     $connection = new mysqli($this->dbhost, $this->dbuser, $this->dbpass, $this->dbname); 
     if($connection->error) 
      die('Could not connect with the database!'); 

     $this->connection = $connection; 
    } 

    public function connect() 
    { 
     $this->__construct($this->dbhost, $this->dbuser, $this->dbpass, $this->dbname); 
    } 

    public function insert($name, $second) 
    { 
     $insert = "INSERT INTO posts (name, second) VALUES ('$name', '$second')"; 

     if ($this->connection->query($insert) === TRUE) 
     { 
      echo "New record created successfully"; 
     } 
     else 
     { 
      echo "Error: " . $sql . "<br>" . mysqli_connect_error(); 
     } 
    } 

    public function getConnection() 
    { 
     return $this->connection; 
    } 
} 

使用

$db = new DB('localhost', 'root', '', 'newdb'); 
$db->insert('name', 'second'); 

我希望这可以帮助你在你的日记，通过OOP，对不起我的英语不好。