Question

我有一个类似于下面的类构造函数（我把一些变量遗漏了），但是我不知道如何构造一个对象并向它发布变量。我继承了这段代码，所以我知道它有效但不确定如何...有人可以告诉我这里发生了什么吗？如何创建对象并设置所有这些参数？

function __construct()
{
$some variables set here 
 $user_name  = isset($_POST['user_name']) ? prepare_input($_POST['user_name'], true) : '';
    $password  = isset($_POST['password']) ? prepare_input($_POST['password'], true) : '';
//some more post variables

$this->wrongLogin = false;      
    if(!$this->IsLoggedIn()){
        if($submit_login == 'login'){
            if(empty($user_name) || empty($password)){
                if(isset($_POST['user_name']) && empty($user_name)){
                    $this->loginError = '_USERNAME_EMPTY_ALERT';                        
                }else if(isset($_POST['password']) && empty($password)){
                    $this->loginError = '_WRONG_LOGIN';
                }
                $this->wrongLogin = true;                           
            }else{
                $this->DoLogin($user_name, $password, $remember_me);
            }
        }else{
            if(isset($_COOKIE[$this->cookieName])){
                parse_str($_COOKIE[$this->cookieName]);
                if(!empty($type) && !empty($usr) && !empty($hash)){
                    $this->accountType = $type;
                    $user_name = $usr;
                    $password = $this->Decrypt($hash, $this->passwordKey);                  
                    $this->DoLogin($user_name, $password, '2');
                }
            }               
        }
    }else if($submit_logout == 'logout'){
        $this->DoLogout();
    }

Answer 1

我不知道这里的问题是什么，但你可以这样做：

class MyClass {
    public $var1;
    public $var2;

    function __construct($var1, $var2 = 5){
        $this->var1 = $var;
        $this->var2 = $var2;
    }
}

代码中的某个地方

// we can skip passing 2nd parameter
$myObject = new MyClass(10);

Answer 2

可以从任何构造函数访问

$ _ POST变量，例如：

<?php
class my_class {
  public $status;      // PROPERTY.
  public $user;        // PROPERTY.
  public $pass;        // PROPERTY.
function __construct ()
{ $this->user = ( isset( $_POST["user"] ) ) ? $_POST["user"] : ""; // PROPERTY.
  $this->pass = ( isset( $_POST["pass"] ) ) ? $_POST["pass"] : ""; // PROPERTY.
  if ( ( strlen( $this->user ) == 0 ) ||
       ( strlen( $this->pass ) == 0 ) )
       $this->status = false;  // PROPERTY.
  else $this->status = true;   // PROPERTY.
}  
}

$myc = new my_class();
$_POST["user"] = "abc"; // HERE $_POST VARIABLES ARE DEFINED AFTER
$_POST["pass"] = "123"; // THE CLASS IS BEEN INSTANTIATED.
if ( $myc->status )
     echo "Good";
else echo "Bad";
?>

它将显示“Bad”，因为实例化类时$ _POST变量不存在，因此构造函数将“”分配给属性。

如何在构造函数中使用post变量？

2 个答案: