Question

我尝试使用此简单代码登录，但是在单击登录按钮后，其他选择为“无效的用户名或密码！”。即使用户名和密码匹配也总是出现

控制器：

class Auth100 extends CI_Controller{

    public function __construct(){
        parent::__construct();
        $this->load->model('AuthModel100');
    }

    public function login(){
        if($this->input->post('login') && $this->validation('login')){
            $login=$this->AuthModel100->getuser($this->input->post('username'));
            if($login !=NULL){
                if(password_verify($this->input->post('password'), $login->password_100)){
                }
            }
            $this->session->set_flashdata('msg','Invalid Username or Password!');
        }
        $this->load->view('auth100/form_login_100');
    }

    public function logout(){
        redirect('auth100/login');
    }

    public function changepass(){
    if($this->input->post('change') && $this->validation('change')){
            redirect('welcome');
        }   
        $this->load->view('auth100/form_password_100');
    }

    public function validation($type){
        $this->load->library('form_validation');    

        if($type=='login'){
            $this->form_validation->set_rules('username', 'Username', 'required');
            $this->form_validation->set_rules('password', 'Password', 'required');
        } else {
            $this->form_validation->set_rules('oldpassword', 'Old Password', 'required');
            $this->form_validation->set_rules('newpassword', 'New Password', 'required');
        }

        if($this->form_validation->run())
            return TRUE;
        else 
            return FALSE;
    }

型号：

class AuthModel100 extends CI_Model{

    public function getuser($username){
        $this->db->where('username_100', $username);
        return $this->db->get('users100')->row();
    }
}

查看：

<html lang="en">
<head>
    <meta charset="UTF-8">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.4.1/css/bootstrap.min.css">
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
    <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.4.1/js/bootstrap.min.js"></script>
    <title>Catshop100 | Login</title>
</head>
<body>
    <div class="container" style="width:50%;">
    <h3>Kitty Fan Shop - Login</h3>
    <hr>
    <div style="color: red;"><?=validation_errors()?></div>
    <form action="" class="form-horizontal" method="post">
        <div class="form-group">
            <label for="username" class="control-label col-sm-2">Username</label>
            <div class="col-sm-10">
                <input type="text" class="form-control" name="username">
            </div>
        </div>
        <div class="form-group">
            <label for="password" class="control-label col-sm-2">Password</label>
            <div class="col-sm-10">
                <input type="password" class="form-control" name="password">
            </div>
        </div>
        <div class="form-group">
            <div class="col-sm-offset-2 col-sm-10">
                <input type="submit" name="login" value="Login" class="btn btn-default">
            </div>
        </div>
    </form>
    </div>
</body>
</html>

数据库：

请帮助我解决此问题

Answer 1

由于错误代码不在else条件下，因此您收到错误消息，我在下面编写了经过纠正的代码，并附有必要的注释。看看是否对您有帮助。

public function login(){

    if($this->input->post('login') && $this->validation('login')){

        $login=$this->AuthModel100->getuser($this->input->post('username'));

        if($login !=NULL){

            if(password_verify($this->input->post('password'), $login->password_100)){ // success
                // do something -- redirect to dashboard -- your logic

            }else{ // fail

                $this->session->set_flashdata('msg','Invalid Username or Password!');

                $this->load->view('auth100/form_login_100');
            }
        }else{

            // do something if $login is NULL

        }
    }
}

Answer 2

尝试一次，您需要处理else条件，以便该方法可以将发生的事情返回给您。

 public function login() {
    if ($this->input->post('login') && $this->validation('login')) {
        $login = $this->AuthModel100->getuser($this->input->post('username'));
        if ($login != NULL) {
            if (password_verify($this->input->post('password'), $login->password_100)) {
                $this->session->set_flashdata('msg', 'Login Successful!');
            } else {
                $this->session->set_flashdata('msg', 'You have entered an invalid password');
            }
        } else {
            $this->session->set_flashdata('msg', 'You have not entered username');
        }
    }
    $this->load->view('auth100/form_login_100');
}

登录表单-无法登录codeigniter

2 个答案: