Question

将数据返回到rest api并且我想返回两个属性和相关属性

这就是我所拥有的

$query = User::find()->joinWith('role');

当我表演时

var_dump($query->all())

我得到一个表格数组（属性和相关属性）

  [1]=>
 object(common\models\User)#119 (10) {
  ["_attributes":"yii\db\BaseActiveRecord":private]=>
array(18) {
  ["id"]=>
  int(27)
  ["username"]=>
  string(8) "marshal2"
  ["email"]=>
  string(18) "marshal2@gmail.com"
 }

["_related":"yii\db\BaseActiveRecord":private]=>
array(1) {
  ["role"]=>
  object(common\rbac\models\Role)#131 (8) {
    ["_attributes":"yii\db\BaseActiveRecord":private]=>
    array(3) {
      ["item_name"]=>
      string(23) "Tracking center officer"
      ["user_id"]=>
      int(27)

    }

  }
}

} ...

通过

将结果传递给json输出时

return ['data' => $query->all()];

仅传递属性但未传递相关属性那是我得到的

"data": [
{
  "username": "admin",
  "email": "track.yard@gmail.com",

},
{
  //also id
  "username": "marshal1",
  "email": "marshal1@gmail.com",
  ..... and others

  //i expected to see a role name since there is a related attribute with user id
},

我需要添加什么才能传递相关属性因为我希望数据也显示角色项名称

这是我的用户模型

    public $role;

public function rules()
{
    return [
        [['username', 'email'], 'filter', 'filter' => 'trim'],
        [['username','expires_on', 'email', 'status'], 'required'],
        ['email', 'email'],
       // ['expires_on', 'integer'],
        [['username','first_name','last_name','firebase_pwd','picture'], 'string', 'min' => 2, 'max' => 255],

        // password field is required on 'create' scenario
        ['password', 'required', 'on' => 'create'],
        // use passwordStrengthRule() method to determine password strength
        $this->passwordStrengthRule(),

        ['username', 'unique', 'message' => 'This username has already been taken.'],
        ['email', 'unique', 'message' => 'This email address has already been taken.'],
    ];
}

   public function fields()
{
    $fields = parent::fields();

    // remove fields that contain sensitive information
    unset($fields['auth_key'], $fields['password_hash'],
        $fields['password_reset_token'],
        $fields['created_at'],
        $fields['firebase_pwd'],
        $fields['updated_at'],
        $fields['id'],
        $fields['authtoken'],
        $fields['account_activation_token']
    );

    $fields['role'] =function ($model){ //added this to see if it'll work
        return $model->role->name;
    };


   return $fields;
}

关系仍在用户模型中

    public function getRole()
{
    // User has_one Role via Role.user_id -> id
    return $this->hasOne(Role::className(), ['user_id' => 'id']);
}

enter code here

Answer 1

在联接上使用eager loading并使用asArray()将数据作为数组返回：

$query = User::find()->with('role')->asArray();

或

$query = User::find()->joinWith('role', true)->asArray();

Answer 2

您可以设置响应i JSON格式

 \Yii::$app->response->format = \yii\web\Response::FORMAT_JSON;
 $items = $query->all();
 return ['data' => $items];

或用php编码

$my_json = json_encode($items);

echo $my_json;

Answer 3

您需要做的就是使用 asArray（）。

示例：

SAFE_DIVIDE(x, y)

Answer 4

您可以接收结果到数组并返回相同的@topher。或者你可以结束阵营。

ArrayHelper::toArray(User::find()->with('role')->all(), [ //or your namespace model 'common\models\User' => ArrayHelper::merge( User::attributes(), ['role' => function($model) { return $model->role; }] ) ])

使用此方法，您可以设置返回结果。

ArrayHelper::toArray(User::find()->all(), [ //or your namespace model 'common\models\User' => [ 'id', 'username' ], ])

Converting Objects to Arrays

P.S这是因为响应准备休息serializer并且它只使用模型中的一个层。

将yii2中的相关属性传递给json输出

4 个答案: