Laravel多对多 - 意外的结果设置 - > select()

Question

我想知道是否有人可以提供帮助,因为我已经撞墙而且还在学习Laravel ORM.我可以解释为什么,当我跑:

public function locationTags(){
    return $this->hasMany('App\UserHasLocationTags', 'user_id')
        ->join('location_tags AS lt', 'lt.id', '=', 'location_tag_id');
}

我得到这个结果集:(剪断......)

{
    "id": 1,
    "created_at": "2015-05-13 13:04:56",
    "updated_at": "2015-05-13 13:04:56",
    "email": "REMOVED",
    "firstname": "REMOVED",
    "lastname": "REMOVED",
    "location_id": 0,
    "deleted_at": null,
    "permissions": [],
    "location_tags": [
        {
            "user_id": 1,
            "location_tag_id": 1,
            "id": 1,
            "created_at": "2015-05-13 13:06:28",
            "updated_at": "2015-05-13 13:06:28",
            "name": "Test Tag 0",
            "location_id": 1,
            "deleted_at": null
        },
        {
            "user_id": 1,
            "location_tag_id": 2,
            "id": 2,
            "created_at": "2015-05-13 11:40:21",
            "updated_at": "2015-05-13 12:56:13",
            "name": "Test Tag 123",
            "location_id": 1,
            "deleted_at": null
        }
    ]
}

哪个是王牌!但是,当我开始从location_tags join中选择我想要的列时,使用:

public function locationTags(){
    return $this->hasMany('App\UserHasLocationTags', 'user_id')
        ->join('location_tags AS lt', 'lt.id', '=', 'location_tag_id')
        ->select('lt.id', 'lt.name');
}

我最终得到:

{
    "id": 1,
    "created_at": "2015-05-13 13:04:56",
    "updated_at": "2015-05-13 13:04:56",
    "email": "REMOVED",
    "firstname": "REMOVED",
    "lastname": "REMOVED",
    "location_id": 0,
    "deleted_at": null,
    "permissions": [],
    "location_tags": []
}

有人可以解释发生了什么吗？并且可能指出我正确的方向来限制选择？谢谢!

更新 我也尝试过:

        $query = \App\User::with(['permissions', 'locationTags' => function($query){
            $query->select('lt.id', 'lt.name');
        }]);

返回相同的结果:(

Answer 1

弄清楚了.这里的关键是你必须包含select()Laravel可用于映射结果集的至少一个键的值.就我而言user_id,就像这样:

public function locationTags(){
    return $this->hasMany('App\UserHasLocationTags', 'user_id')
        ->join('location_tags AS lt', 'lt.id', '=', 'location_tag_id')
        ->select('user_id', 'lt.name', 'location_tag_id');
}

然后返回更好的结果集:

{
    "id": 1,
    "created_at": "2015-05-13 13:04:56",
    "updated_at": "2015-05-13 13:04:56",
    "email": "REMOVED",
    "firstname": "REMOVED",
    "lastname": "REMOVED",
    "location_id": 0,
    "deleted_at": null,
    "permissions": [],
    "location_tags": [
        {
            "user_id": 1,
            "name": "Test Tag 0",
            "location_tag_id": 1
        },
        {
            "user_id": 1,
            "name": "Test Tag 123",
            "location_tag_id": 2
        }
    ]
}

希望这可以帮助将来的某个人,因为它让我猜了好几个小时.