我想顯示當(dāng)前用戶的姓名。在視圖中 (dashboard.php)<?php $username = $this->session->userdata('logged_in');echo $username;?>我的數(shù)據(jù)庫(kù)在表“用戶”中有一個(gè)字段“名稱”。用戶名保存在這里。但我得到的輸出只是“名稱”所以，我在下面添加了這個(gè)<?php $username = $this->session->userdata('logged_in');echo $username;print_r($this->session->all_userdata());?>輸出是名稱數(shù)組（[__ci_last_regenerate] => 1583656504 [name] => name [] => [logged_in] => 1）問題在哪里？我注冊(cè)了新用戶，值已正確保存在數(shù)據(jù)庫(kù)中，我也可以登錄。但無法顯示登錄用戶名。Login.php（控制器）<?phpdefined('BASEPATH') OR exit('No direct script access allowed');class Login extends CI_Controller {function __construct() {parent::__construct();// Load url helper $this->load->helper('url'); $this->load->library(['form_validation','session']); $this->load->database(); //load the Login Model $this->load->model('UserLogin', 'login');} public function index() { $logged_in = $this->session->userdata('logged_in'); if($logged_in){ //if yes redirect to welcome page redirect(base_url().'app/dashboard'); } $data['title'] = 'Login'; $this->load->view('app/login', $data); } public function doLogin() { //get the input fields from login form $name = $this->input->post('name'); $email = $this->input->post('email'); $password = sha1($this->input->post('password')); //send the email pass to query if the user is present or not $check_login = $this->login->checkLogin($email, $password); //if the result is query result is 1 then valid user if ($check_login) { // $this->session->set_userdata('logged_in', true); $this->session->set_userdata('logged_in', $check_login); redirect(base_url().'app/dashboard'); }}
3 回答
嚕嚕噠
TA貢獻(xiàn)1784條經(jīng)驗(yàn) 獲得超7個(gè)贊
您的方法有幾個(gè)問題，您正在打印用戶名
$username = $this->session->userdata('logged_in');
echo $username;
但是您沒有在任何地方分配用戶名，因此要完成此操作，您可以在登錄類中進(jìn)行一些選擇
class UserLogin extends CI_Model{
public function checkLogin($email, $password) {
$this->db->where('email', $email);
$this->db->where('password', $password);
$query = $this->db->get('users');
if($query->num_rows())
{
$user = $query->row_array();
print_r($user); die(); // remove die after testing
return $user;
}
return false;
}
比你能回聲：
$user = $this->session->userdata('logged_in');
echo $user['username'];
第二個(gè)問題是這種方式登錄用戶非常不安全，您可以對(duì)密碼進(jìn)行md5或其他加密。IE：
$this->db->where('password', md5( $password));
或更好的用戶ion_auth library，默認(rèn)情況下為您提供各種選項(xiàng)。
ibeautiful
TA貢獻(xiàn)1993條經(jīng)驗(yàn) 獲得超6個(gè)贊
更改以下CheckLogin()類似然后檢查用戶數(shù)據(jù)
public function checkLogin($email, $password) {
$this->db->where('email', $email);
$this->db->where('password', $password);
$query = $this->db->get('users');
if($query->num_rows() == '1'){
$data = $query->row_array();
$this->session->set_userdata(array('name'=>$data['name'],'loggedin'=>'1'));
}
return $query->num_rows();
米琪卡哇伊
TA貢獻(xiàn)1998條經(jīng)驗(yàn) 獲得超6個(gè)贊
請(qǐng)嘗試更改模型
class UserLogin extends CI_Model
{
public function checkLogin($email, $password) {
$this->db->where('email', $email);
$this->db->where('password', $password);
return $this->db->get('users');
}
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