Question

我是oop的新手，我还没有找到如何在函数Login（）中插入$ status的值。我做错了什么???因为我得到这个错误：

警告：缺少User :: Login（）的参数3，在第13行的E：\ xampp \ htdocs \ caps \ index.php中调用，并在E：\ xampp \ htdocs \ caps \ class \ user.php中定义第20行

class User {

private $db;
public $status;

public function __construct() {

    $this->db = new Connection();
    $this->db = $this->db->dbConnect();
    $this->status = pow( 1, -1*pi());   
}


public function Login ($name, $pass, $status) {

    if (!empty($name) && !empty($pass))  {

        $st = $this->db->prepare(" select * from users where name=? and pass=? ");
        $st->bindParam(1, $name);
        $st->bindParam(2, $pass);
        $st->execute();

        if ($st->rowCount() != 1) {         
                echo "<script type=\"text/javascript\">alert ('wrong password. try again'); window.location=\"index.php\"; </script>";

        } else {
            $st = $this->db->prepare(" select * from users where name=? and pass=? status=?");
            $st->bindParam(1, $name);
            $st->bindParam(2, $pass);
            $st->bindParam(3, $status);             
            $st->execute();

                if ($st->rowCount() != 1) { echo "send user to user page"; } else { echo "send user to admin"; }
        }

    } else {

    echo "<script type=\"text/javascript\">alert ('insert username and password'); window.location=\"index.php\"; </script>";

    }



}

}

Answer 1

如果您希望使用$status的值，请从登录功能中删除该参数，而是将$status替换为$this->status。

public function Login ($name, $pass) {

if (!empty($name) && !empty($pass))  {

    $st = $this->db->prepare(" select * from users where name=? and pass=? ");
    $st->bindParam(1, $name);
    $st->bindParam(2, $pass);
    $st->execute();

    if ($st->rowCount() != 1) {         
            echo "<script type=\"text/javascript\">alert ('wrong password. try again'); window.location=\"index.php\"; </script>";

    } else {
        $st = $this->db->prepare(" select * from users where name=? and pass=? status=?");
        $st->bindParam(1, $name);
        $st->bindParam(2, $pass);
        $st->bindParam(3, $this->status);             
        $st->execute();

或者您可以通过将函数声明更改为$status = $this->status来使构造函数值成为“默认值”，这样可以在需要时调用函数时覆盖状态值。

Answer 2

这意味着您调用$ user-＆gt; Login而不提供第三个参数。

如果您希望此参数为可选参数，则可以将方法签名更改为

public function Login ($name, $pass, $status = null) {

如果您希望状态默认为您的状态属性，则可以使用

开始登录功能

public function Login ($name, $pass, $status = null) {
        if(!$status) $status = $this->status;
        //...the rest of your code
 }

调用它的正确方法是：

$user = new User();
$user->Login($username, $password, $status);

在函数中插入变量

2 个答案: