4
我是新來的Yii和我有問題，它的用戶和登錄系統Yii2用戶模型。 我應該使用登錄檢查3個表，但是當我使用自定義查詢我面對與關係表
"Argument 1 passed to yii\web\User::login() must implement interface yii\web\IdentityInterface, ActiveQuery given"
我的表是這樣的：
user : user_id, name, family, birthday, ...email : email_user_fk, email_addr, email_active, email_cdtpasswd : passwd_user_fk, passwd_hashed, passwd_active, passwd_cdt
和我的查詢是這樣的： SELECT user.user_id, email.email_addr, email.email_active, passwd.passwd_hashed, passwd_passwd_active , ... FROM user JOIN email ON user.user_id = email.email_user_fk JOIN passwd ON user.user_id = passwd.passwd_user_fk WHERE email.email_addr = :email
有什麼想法嗎？
class User extends ActiveRecord implements IdentityInterface
{
const STATUS_DELETED = 0;
const STATUS_ACTIVE = 10;
/**
* @inheritdoc
*/
public static function tableName()
{
return 'user';
}
public static function primaryKey(){
return 'user_id';
}
/**
* @inheritdoc
*/
public function behaviors()
{
return [
TimestampBehavior::className(),
];
}
/**
* @inheritdoc
*/
public static function find
Identity($id)
{
return static::findOne(['id' => $id, 'status' => self::STATUS_ACTIVE]);
}
/**
* @inheritdoc
*/
public static function findIdentityByAccessToken($token, $type = null)
{
throw new NotSupportedException('"findIdentityByAccessToken" is not implemented.');
}
/**
* Finds user by username
*
* @param string $username
* @return static|null
*/
public static function findByUsername($username)
{
return static::findOne(['email' => $username]);
}
public static function findByEmail($email)
{
return User::find()
->joinWith(['emails'])
->where("email.email_address = '[email protected]' ")
->one();
}
public static function findByMobile($email)
{
return User::find()
->joinWith(['mobiles'])
->where("mobile.mobile_address = '0931515124' ")
->one();
}
/**
* Finds user by password reset token
*
* @param string $token password reset token
* @return static|null
*/
public static function findByPasswordResetToken($token)
{
if (!static::isPasswordResetTokenValid($token)) {
return null;
}
return static::findOne([
'password_reset_token' => $token,
'status' => self::STATUS_ACTIVE,
]);
}
/**
* Finds out if password reset token is valid
*
* @param string $token password reset token
* @return boolean
*/
public static function isPasswordResetTokenValid($token)
{
if (empty($token)) {
return false;
}
$expire = Yii::$app->params['user.passwordResetTokenExpire'];
$parts = explode('_', $token);
$timestamp = (int) end($parts);
return $timestamp + $expire >= time();
}
/**
* @inheritdoc
*/
public function getId()
{
return $this->getPrimaryKey();
}
/**
* @inheritdoc
*/
public function getAuthKey()
{
return $this->auth_key;
}
/**
* @inheritdoc
*/
public function validateAuthKey($authKey)
{
return $this->getAuthKey() === $authKey;
}
/**
* Validates password
*
* @param string $password password to validate
* @return boolean if password provided is valid for current user
*/
public function validatePassword($password)
{
return Yii::$app->security->validatePassword($password, $this->password_hash);
}
/**
* Generates password hash from password and sets it to the model
*
* @param string $password
*/
public function setPassword($password)
{
$this->password_hash = Yii::$app->security->generatePasswordHash($password);
}
/**
* Generates "remember me" authentication key
*/
public function generateAuthKey()
{
$this->auth_key = Yii::$app->security->generateRandomString();
}
/**
* Generates new password reset token
*/
public function generatePasswordResetToken()
{
$this->password_reset_token = Yii::$app->security->generateRandomString() . '_' . time();
}
/**
* Removes password reset token
*/
public function removePasswordResetToken()
{
$this->password_reset_token = null;
}
public function rules()
{
return [
[['user_name', 'user_family', 'user_birthday'], 'required'],
[['user_gender', 'city_id_fk', 'user_status'], 'integer'],
[['user_birthday', 'user_cdt'], 'safe'],
[['user_name'], 'string', 'max' => 32],
[['user_family'], 'string', 'max' => 48],
[['user_tel', 'user_postcode'], 'string', 'max' => 12],
[['user_address'], 'string', 'max' => 128],
[['user_profile_image', 'user_cover_image'], 'string', 'max' => 256]
];
}
/**
* @return \yii\db\ActiveQuery
*/
public function getEmails()
{
return $this->hasMany(Email::className(), ['email_user_id_fk' => 'user_id']);
}
/**
* @return \yii\db\ActiveQuery
*/
public function getMobiles()
{
return $this->hasMany(Mobile::className(), ['mobile_user_id_fk' => 'user_id']);
}
/**
* @return \yii\db\ActiveQuery
*/
public function getPasswds()
{
return $this->hasMany(Passwd::className(), ['passwd_user_id_fk' => 'user_id']);
}
}
顯示您的用戶模式或其中定義的login（）方法 – scaisEdge
@scaisEdge模型 - 比如我要定義的函數findByEmail（），但電子郵件列不會在用戶表存儲。電子郵件已將表格與一對多關係分開。我需要加入用戶表和電子郵件表，其中email_active爲true – masoud
在你的代碼中，你似乎沒有正確實現'user interface'。似乎沒有'findIndentity（）'方法。接口的實現必須實現所有的接口方法。 – scaisEdge