Question

将$ con转移到类中的方法上，我必须更改什么？

它应该是“ public”之类的。

class MyClass {

public function __construct()
{
    $con=mysqli_connect("localhost", "dbu...", "pw") or die("Error");
    mysqli_select_db($con, "db100...") or die("Error");
}

public function doSomething()
{
    mysqli_query($con, "");
}

}

谢谢！

PS：我不要$ this-> db-> query（）或其他东西：）

Answer 1

如果要坚持使用mysqli的过程版本，则在面向对象的类中，仍然需要使用$this，但要有所不同：

class MyClass {
    private $con;
    public function __construct()
    {
        $this->con = mysqli_connect("localhost", "dbu...", "pw") or die("Error");
        mysqli_select_db($this->con, "db100...") or die("Error");
    }

    public function doSomething()
    {
        mysqli_query($this->con, "");
    }
}

但是理想情况下，如果您坚持使用对象表示法，则应该一直使用object oriented：

class MyClass {
    private $con;
    public function __construct()
    {
        $this->con = new mysqli("localhost", "dbu...", "pw") or die("Error");
        $this->con->select_db("db100...") or die("Error");
    }

    public function doSomething()
    {
        $this->con->query("");
    }
}

Answer 2

您可以这样做：

class MyClass {

  public $con

  public function __construct()
  {
    $this->con=mysqli_connect("localhost", "dbu...", "pw") or die("Error");
    mysqli_select_db($this->con, "db100...") or die("Error");
  }

  public function doSomething()
  {
    mysqli_query($this->con, "");
  }

}

...或这样：

global $con; 

class MyClass {

  public function __construct()
  {
    global $con;
    $con=mysqli_connect("localhost", "dbu...", "pw") or die("Error");
    mysqli_select_db($con, "db100...") or die("Error");
  }

  public function doSomething()
  {
    global $con;
    mysqli_query($con, "");
  }

}

建议使用第一个选项，因为第二个选项可能导致名称空间冲突

PHP：MySQLi Connection在方法中可用

2 个答案: