我想实现一个使用OpenID的登录方法,并且$ _SESSION var的正确发布 - 通过那些我只是试图在MySQL中注册用户。出于某种原因,穿过我的控制器::Yii2 save()使用默认值创建数据库行
public function actionLogin()
{
if (!Yii::$app->user->isGuest) {
return $this->goHome();
}
include ('../views/user-record/steamauth/userInfo.php');
$steamid = $_SESSION['steam_steamid'];
$username = $_SESSION['steam_personaname'];
$profileurl = $_SESSION['steam_profileurl'];
$avatar = $_SESSION['steam_avatar'];
$avatarmedium = $_SESSION['steam_avatarmedium'];
$avatarfull = $_SESSION['steam_avatarfull'];
$user = UserRecord::findBySteamId($steamid);
if ($user === null)
{
$user = new UserRecord();
$user->steamid = $steamid;
$user->username = $username;
$user->profileurl = $profileurl;
$user->avatar = $avatar;
$user->avatarmedium = $avatarmedium;
$user->avatarfull = $avatarfull;
$user->verified = 0;
$user->banned = 0;
$user->save();
}
Yii::$app->user->login($user, 604800);
return $this->redirect(Yii::$app->user->returnUrl);
}
编辑“登录”操作时:这里是UserRecord类,忘了将它添加在
<?php
namespace app\models;
class UserRecord extends \yii\db\ActiveRecord implements \yii\web\IdentityInterface
{
public $id;
public $steamid;
public $username;
public $profileurl;
public $avatar;
public $avatarmedium;
public $avatarfull;
public $verified;
public $banned;
public $rank;
public $authKey;
// public $password;
// public $accessToken;
public static function tableName()
{
return 'users';
}
public function getAuthKey()
{
return $this->authKey;
}
public function getId()
{
return $this->id;
}
public function validateAuthKey($authKey)
{
return $this->authKey === $authKey;
}
public static function findIdentity($id)
{
return self::findOne($id);
}
public function validateSteamID($steamid)
{
return $this->steamid === $steamid;
}
public static function findIdentityByAccessToken($token, $type = null)
{
throw new \yii\base\NotSupportedException;
}
public static function findBySteamId($steamid)
{
return self::findOne(['steamid' => $steamid]);
}
}
结果是一个简单的。张贴行,没有输入数据。
任何帮助将不胜感激,谢谢。
你正在使用的定义在你的模态属性数组中的? – Dani
更新了@Dani谢谢 –