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yii2 如何處理管理員登陸

yii2 如何處理管理員登陸

繁華開滿天機 2019-04-13 08:45:46
提交的數(shù)據(jù)$_GET=['r'=>'site/login',];$_POST=['_csrf'=>'WjgxM0tfa3YxQn1rcg4vHioLVlt.EjhOAnFVR3MuXgQ7WkNmBwofIg==','LoginForm'=>['username'=>'admin','password'=>'admin','rememberMe'=>'1',],'login-button'=>'',];yii2處理管理員登陸部分的代碼:#/var/www/example.com/public_html/yii/backend/controllers/SiteController.phppublicfunctionactionLogin(){if(!\Yii::$app->user->isGuest){return$this->goHome();}$model=newLoginForm();if($model->load(Yii::$app->request->post())&&$model->login()){return$this->goBack();}else{return$this->render('login',['model'=>$model,]);}}但是我并沒有user表,這里是backend,管理員登陸部分,對應的表是admin,結構。mysql>descadmin;+-------+------------------+------+-----+---------+-------+|Field|Type|Null|Key|Default|Extra|+-------+------------------+------+-----+---------+-------+|id|int(10)unsigned|NO||NULL|||name|char(30)|NO|||||pwd|char(32)|NO|||||email|char(30)|NO||||+-------+------------------+------+-----+---------+-------+4rowsinset(0.03sec)mysql>select*fromadmin;+----+---------+----------------------------------+-------+|id|name|pwd|email|+----+---------+----------------------------------+-------+|1|admin|b9840270f7f4ed699ccc93cc09914d21||+----+---------+----------------------------------+-------+1rowinset(0.00sec)yii2他默認的都是user表,可我并沒有user表。我現(xiàn)在提交了用戶名和密碼,該如何修改actionLogin里面的代碼才最合適。
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TA貢獻1775條經驗 獲得超8個贊

試了一下2.0,把models/User.php修改成下邊這樣.
namespaceapp\models;
//繼承關系修改為如下
classUserextends\yii\db\ActiveRecordimplements\yii\web\IdentityInterface
{
public$id;
//username字段修改為name
public$name;
//password字段修改為pwd
public$pwd;
public$authKey;
public$accessToken;
publicstaticfunctiontableName()
{
//數(shù)據(jù)庫表名修改user為admin
return'{{admin}}';
}
publicstaticfunctionfindIdentity($id)
{
returnstatic::findOne($id);
}
publicstaticfunctionfindIdentityByAccessToken($token)
{
returnstatic::findOne(['access_token'=>$token]);
}
publicstaticfunctionfindByUsername($username)
{
//username修改為name
returnstatic::findOne(['name'=>$username]);
}
publicfunctiongetId()
{
return$this->id;
}
publicfunctiongetAuthKey()
{
return$this->authKey;
}
publicfunctionvalidateAuthKey($authKey)
{
return$this->authKey===$authKey;
}
publicfunctionvalidatePassword($password)
{
//password修改為pwd
return$this->pwd===$password;
}
}
                            
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反對 回復 2019-04-13
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