Question

我正在尝试获取具有关系和嵌套字段的用户模型。第一次尝试运行良好：

$user = \App\User::where('id', auth()->user()->id)->with(['userdoc'])->get();

它返回整个userdoc对象。但是我只想要两个字段，所以我在尝试：

$user = \App\User::where('id', auth()->user()->id)->with(['userdoc:id,number'])->get();

通过这种方式，它以userdoc:null JSON返回toArray()。

根据https://laravel.com/docs/5.5/eloquent-relationships#eager-loading上的文档，这似乎没问题

我正在使用Laravel 5.6。

添加更多信息：

没有特定字段将获取整个对象

$user = \App\User::where('id', auth()->user()->id)->with('userdoc')->get();
return response()->json($user->toArray(), 200);

[
    {
        "id": 11,
        "email": "xxxxx",
        "slug": "xxxxx",
        "created_at": "xxxxx",
        "updated_at": "xxxxx",
        "deleted_at": null,
        "first_name": "xxxxx",
        "last_name": "xxxxx",
        "is_verified": 0,
        "userdoc": {
            "id": 11,
            "created_at": "2018-08-20 13:20:29",
            "updated_at": "2018-08-20 13:20:29",
            "deleted_at": null,
            "user_id": 11,
            "type": "document",
            "number": "xxxxxxxxx"
        }
    }
]

确定userdoc字段会使对象变为空

$user = \App\User::where('id', auth()->user()->id)->with('userdoc:id,number')->get();
return response()->json($user->toArray(), 200);

[
    {
        "id": 11,
        "email": "xxxxxxxxxxx",
        "slug": "xxxxxxxxxxx",
        "created_at": "xxxxxxxxxxx",
        "updated_at": "xxxxxxxxxxx",
        "deleted_at": null,
        "first_name": "xxxxxxxxxxx",
        "last_name": "xxxxxxxxxxx",
        "is_verified": 0,
        "userdoc": null
    }
]

Answer 1

您需要在以下位置删除括号=Parameters!Deliverable.Value <> "ABC" AND Parameters!Deliverable.Value <> "All"：

[]

应该是：

$user = \App\User::where('id', auth()->user()->id)->with(['userdoc:id,number'])->get();
_________________________________________________________^___________________^

您也应该在$user = \App\User::where('id', auth()->user()->id)->with('userdoc:id,number')->get();传递的字段中传递外键：

with

Answer 2

您得到的结果是因为您没有选择相关的主键user_id。试试这个

$user = \App\User::where('id', auth()->user()->id)->with(['userdoc:id,user_id,number'])->get();

Answer 3

我猜测您的userdocs表具有user_id属性，并且您正在users和userdocs之间使用一对多关系。在这种情况下，您还需要从userdoc获取外键：

$user = \App\User::where('id', auth()->id())->with(['userdoc:id,number,user_id'])->get();

Answer 4

我必须将id添加到嵌套的紧急加载中，以获取客户端的名称：

$foo = Visitor_visit::with('client:id,name')->get();

否则，它为空，希望能对某人有所帮助。

雄辩的急切加载不适用于嵌套字段

4 个答案: