我尝试在Yii基本模板中实现User Model和LoginForm Model来验证用户登录。我创建了一个数据库并连接到它。作为表用户的数据库和名为username，password，authKey和acessToken的字段填充了值。从ActiveRecord扩展用户模型并实现\ yii \ web \ IdentityInterface，以使内置的Yii2功能完成其工作。还写了这个方法：
public static function tableName() { return 'user'; }
每次我尝试登录时都会抛出 - ＆gt;用户名或密码不正确，来自LoginForm模型中的validatepassword()。
这是我的代码：
LoginForm模型：
<?php
namespace app\models;
use Yii;
use yii\base\Model;
/**
* LoginForm is the model behind the login form.
*/
class LoginForm extends Model
{
public $username;
public $password;
public $rememberMe = true;
private $_user = false;
/**
* @return array the validation rules.
*/
public function rules()
{
return [
// username and password are both required
[['username', 'password'], 'required'],
// rememberMe must be a boolean value
['rememberMe', 'boolean'],
// password is validated by validatePassword()
['password', 'validatePassword'],
];
}
/**
* Validates the password.
* This method serves as the inline validation for password.
*
* @param string $attribute the attribute currently being validated
* @param array $params the additional name-value pairs given in the rule
*/
public function validatePassword($attribute, $params)
{
if (!$this->hasErrors()) {
$user = $this->getUser();
if (!$user || !$user->validatePassword($this->password)) {
$this->addError($attribute, 'Incorrect username or password.');
}
}
}
/**
* Logs in a user using the provided username and password.
* @return boolean whether the user is logged in successfully
*/
public function login()
{
if ($this->validate()) {
return Yii::$app->user->login($this->getUser(), $this->rememberMe ? 3600*24*30 : 0);
} else {
return false;
}
}
/**
* Finds user by [[username]]
*
* @return User|null
*/
public function getUser()
{
if ($this->_user === false) {
$this->_user = User::findByUsername($this->username);
}
return $this->_user;
}
}
...这是我的User.php模型：
<?php
namespace app\models;
use yii\db\ActiveRecord;
class User extends ActiveRecord implements \yii\web\IdentityInterface
{
public $id;
public $username;
public $password;
public $authKey;
public $accessToken;
public static function tableName() { return 'user'; }
/**
* @inheritdoc
*/
public static function findIdentity($id) {
$user = self::find()
->where([
"id" => $id
])
->one();
if (!count($user)) {
return null;
}
return new static($user);
}
/**
* @inheritdoc
*/
public static function findIdentityByAccessToken($token, $userType = null) {
$user = self::find()
->where(["accessToken" => $token])
->one();
if (!count($user)) {
return null;
}
return new static($user);
}
/**
* Finds user by username
*
* @param string $username
* @return static|null
*/
public static function findByUsername($username) {
$user = self::find()
->where([
"username" => $username
])
->one();
if (!count($user)) {
return null;
}
return new static($user);
}
/**
* @inheritdoc
*/
public function getId() {
return $this->id;
}
/**
* @inheritdoc
*/
public function getAuthKey() {
return $this->authKey;
}
/**
* @inheritdoc
*/
public function validateAuthKey($authKey) {
return $this->authKey === $authKey;
}
/**
* Validates password
*
* @param string $password password to validate
* @return boolean if password provided is valid for current user
*/
public function validatePassword($password) {
return $this->password === $password;
}
}
我不知道我该怎么办，也许在验证密码或查找用户名方面存在问题，在Yii2调试中它显示正确连接到mysql数据库。
不要使用siteController actionLogin()，因为它等同于高级模板，我认为保持这种方式是正确的。
简而言之，以下功能不断抛出＆＃34;不正确的用户名或密码。＆＃34;：
public function validatePassword($attribute, $params)
{
if (!$this->hasErrors()) {
$user = $this->getUser();
if (!$user || !$user->validatePassword($this->password)) {
$this->addError($attribute, 'Incorrect username or password.');
}
}
}
我不想放弃，但我考虑回到旧的Yii1.xx.在那里，我可以轻松地查询数据库并使一个良好的登录系统正常工作。
我花了将近72个小时来解决此登录问题，并且没有解决方案解决了基本的Yii2模板。
我不想使用默认包中的静态$users。
编辑2
的 siteController.php
<?php
namespace app\controllers;
use Yii;
use yii\filters\AccessControl;
use yii\web\Controller;
use yii\filters\VerbFilter;
use app\models\LoginForm;
use app\models\ContactForm;
use yii\helpers\url;
class SiteController extends Controller
{
public function behaviors()
{
return [
'access' => [
'class' => AccessControl::className(),
'only' => ['logout'],
'rules' => [
[
'actions' => ['logout'],
'allow' => true,
'roles' => ['@'],
],
],
],
'verbs' => [
'class' => VerbFilter::className(),
'actions' => [
'logout' => ['post'],
],
],
];
}
public function actions()
{
return [
'error' => [
'class' => 'yii\web\ErrorAction',
],
'captcha' => [
'class' => 'yii\captcha\CaptchaAction',
'fixedVerifyCode' => YII_ENV_TEST ? 'testme' : null,
],
];
}
public function actionIndex()
{
return $this->render('index');
}
public function actionLogin()
{
if (!\Yii::$app->user->isGuest) {
return $this->goHome();
}
$model = new LoginForm();
if ($model->load(Yii::$app->request->post()) && $model->login()) {
return $this->redirect(Url::toRoute(['contacto/index']));
} else {
return $this->render('login', [
'model' => $model,
]);
}
}
public function actionLogout()
{
Yii::$app->user->logout();
return $this->goHome();
}
public function actionContact()
{
$model = new ContactForm();
if ($model->load(Yii::$app->request->post()) && $model->contact(Yii::$app->params['adminEmail'])) {
Yii::$app->session->setFlash('contactFormSubmitted');
return $this->refresh();
} else {
return $this->redirect(Url::toRoute(['contacto/create2']));
}
}
public function actionAbout()
{
return $this->render('about');
}
public function actionSkills()
{
return $this->render('skills');
}
public function actionPortfolio()
{
return $this->render('portfolio');
}
// tradução do site
public function beforeAction($action) {
if (Yii::$app->session->has('lang')) {
Yii::$app->language = Yii::$app->session->get('lang');
} else {
Yii::$app->language = 'us';
}
return parent::beforeAction($action);
}
public function actionLangus(){
Yii::$app->session->set('lang', 'us'); //or $_GET['lang']
return $this->redirect(Url::toRoute(['site/index']));
}
public function actionLangpt(){
Yii::$app->session->set('lang', 'pt'); //or $_GET['lang']
return $this->redirect(Url::toRoute(['site/index']));
}
}
答案 0 :(得分：2)
1）停止创建重复的问题 2）删除变量的公共声明。
class User extends ActiveRecord implements \yii\web\IdentityInterface
{
public $id;
public $username;
public $password;
public $authKey;
public $accessToken;
$ user-＆gt;密码将为空。而不是使用魔术方法来获取实际上声明它们的值......当你使用它们时它们将永远是空的。