Question

实际上，我想使用Model对象而不是通过下面提到的Model Facade来获取数据。

$user = $this->getGuard()->user();

以上用户是当前登录的用户，

        "id": 6,
        "name": "kuser",
        "email": "kuser@gmail.com",
        "phone": "03345154067",
        "is_subscribed": 0,
        "subscription_date": null

但是当我尝试使用其相关模型（如卡）来获取相同的$user对象时，通过执行$user->with(['cards'])->first();会比使用如下所述的卡给出用户表的第一条记录。

        "id": 1,
        "name": "admin",
        "email": "admin@gmail.com",
        "phone": "03056494616",
        "is_subscribed": 1,
        "subscription_date": "2019-10-08",
        "cards": []

上面提到的记录是用户对象。这是我的用户表的第一条记录。

实际上，我期待。

      "id": 6,
        "name": "kuser",
        "email": "kuser@gmail.com",
        "phone": "03345154067",
        "is_subscribed": 0,
        "subscription_date": null

及其相关的模型卡，例如

      "id": 6,
        "name": "kuser",
        "email": "kuser@gmail.com",
        "phone": "03345154067",
        "is_subscribed": 0,
        "subscription_date": null
        "cards":[] // array of cards those belongs to it.

Answer 1

您可以使用Lazy Eager Loading，而不必再次查询用户，您已经拥有了：

$user = $this->getGuard()->user();

$user->load('cards');

Laravel 6 Docs - Eloquent - Relationships - Lazy Eager Loading

Answer 2

使用简单的public class SchoolsController : ODataController { private readonly ISchoolService _schoolService; public SchoolsController(ISchoolService schoolService) { _schoolService = schoolService; } [EnableQuery(PageSize = 10,MaxExpansionDepth= 12,MaxAnyAllExpressionDepth = 5)] public IQueryable<School> Get() { return _schoolService.Schools(); } [HttpGet] [EnableQuery] public IQueryable<School> Schools() { return _schoolService.Schools(); } }吗？

where

那行不通吗？

Answer 3

这是另一种方式

User::with(['cards'])->find($user->id);

如何在Laravel中雄伟地使用via对象获取数据？

3 个答案: