yii2.0用户登陆
在yii2的basic版本中默认是从一个数组验证用户名和密码，如何改为从数据表中查询验证呢？
首先新建立一个user表并插入实验数据，如下
CREATE TABLE `user` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`username` varchar(50) NOT NULL,
`password` varchar(32) NOT NULL,
`authKey` varchar(100) NOT NULL DEFAULT '',
`accessToken` varchar(100) NOT NULL DEFAULT '',
PRIMARY KEY (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=3 DEFAULT CHARSET=utf8;
-- ----------------------------
-- Records of myuser
-- ----------------------------
INSERT INTO `user` VALUES ('1', 'admin', '123', '1234', '1234');
INSERT INTO `user` VALUES ('2', 'demo', '234', '123', '123');
模型User.php如下，替换掉原来的User.php就可以实现数据表验证登陆了
namespace app\models;class User extends /*\yii\base\Object*/ \yii\db\ActiveRecord implements \yii\web\IdentityInterface{ /*public $id;
public $username;
public $password;
public $authKey;
public $accessToken;
private static $users = [
'100' => [
'id' => '100',
'username' => 'admin',
'password' => 'admin',
'authKey' => 'test100key',
'accessToken' => '100-token',
],
'101' => [
'id' => '101',
'username' => 'demo',
'password' => 'demo',
'authKey' => 'test101key',
'accessToken' => '101-token',
],
];
*/
/**
* @inheritdoc
*/
public static function tableName()
{ return 'user';
} /**
* @inheritdoc
*/
public function rules()
{ return [
[['username', 'password'], 'required'],
[['username'], 'string', 'max' => 50],
[['password'], 'string', 'max' => 32],
[['authKey'], 'string', 'max' => 100],
[['accessToken'], 'string', 'max' => 100],
];
} /**
* @inheritdoc
*/
public function attributeLabels()
{ return [ 'id' => 'ID', 'username' => 'Username', 'password' => 'Password', 'authKey' => 'AuthKey', 'accessToken' => 'AccessToken',
];
} /**
* @inheritdoc
*/
public static function findIdentity($id)
{ return static::findOne($id); //return isset(self::$users[$id]) ? new static(self::$users[$id]) : null;
} /**
* @inheritdoc
*/
public static function findIdentityByAccessToken($token, $type = null)
{ return static::findOne(['access_token' => $token]); /*foreach (self::$users as $user) {
if ($user['accessToken'] === $token) {
return new static($user);
}
}
return null;*/
} /**
* Finds user by username
*
* @param string $username
* @return static|null
*/
public static function findByUsername($username)
{ $user = User::find()
->where(['username' => $username])
->asArray()
->one(); if($user){ return new static($user);
} return null; /*foreach (self::$users as $user) {
if (strcasecmp($user['username'], $username) === 0) {
return new static($user);
}
}
return null;*/
} /**
* @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 === $password;
}
}
原来User.php是继承了\yii\base\Object，为什么要继承这个类，是有原因的
在\yii\base\Object中，有构造方法 public function __construct($config = [])
{ if (!empty($config)) {
Yii::configure($this, $config);
} $this->init();
}
继续追踪Yii::configure($this, $config)代码如下
public static function configure($object, $properties)
{ foreach ($properties as $name => $value) { $object->$name = $value;
} return $object;
}
正是因为有这两个方法，所以在User.php中public static function findByUsername($username)
{ foreach (self::$users as $user) { if (strcasecmp($user['username'], $username) === 0) { return new static($user);
}
} return null;
}
将$user传递过来，通过static，返回一个User的实例。
当通过数据表查询时候没有必要再继承\yii\base\Object，因为不必为类似原来类变量赋值了。这个时候需要User.php继承\yii\db\ActiveRecord，因为要查询用。
findIdentity是根据传递的id返回对应的用户信息，getId返回用户id，getAuthKey和validateAuthKey是 作用于登陆中的--记住我。这个authKey是唯一的，当再次登陆时，从cookie中获取authKey传递给validateAuthKey，验证 通过，就登陆成功。