Android Room 一对多关系查询反向

Question

我有两个对象之间的一对多关系。
我有一个 user有 many pets .

使用它我可以检索我的对象

data class UserWithPets (
        @Embedded
        var user: User? = null,

        @Relation(parentColumn = "id", entityColumn = "user_id", entity = Pet::class)
        var pets: List<Pet>? = null
)

用户道:

@Transaction @Query("SELECT * FROM users")
fun getUserWithPets() : LiveData<List<UserWithPets>>

这工作正常。
现在我想获取与每个用户相关联的宠物列表......这将是 PetWithUser .

所以我做了:

data class PetWithUser (
        @Embedded
        var pet: Pet? = null,

        @Relation(parentColumn = "user_id", entityColumn = "id", entity = Pet::class)
        var user: User? = null
)

宠物道:

@Transaction @Query("SELECT * FROM pets")
fun getPetsWithUser(): LiveData<List<PetWithUser>>

只要我添加 getPetsWithUser对于代码，我在生成 DataBinding 类时遇到错误。

如何建立一对多关系并获得 pet与 user ?

编辑:

错误日志:它们是从 DataBinding java 文件中生成的。第二个(HomeFragmentDataBinding)不使用 PetWithUser但在 PetWithUser 时引发错误已添加到代码中...

error: cannot find symbol
  protected ListItemBinding(DataBindingComponent _bindingComponent, View _root,
                                   ^
  symbol:   class DataBindingComponent
  location: class ListItemBinding

error: cannot find symbol
      @Nullable ViewGroup root, boolean attachToRoot, @Nullable DataBindingComponent component) {
                                                                ^
  symbol:   class DataBindingComponent
  location: class FragmentHomeBinding

Answer 1

正确的版本是:

data class PetWithUser (
    @Embedded
    var pet: Pet,

    @Relation(parentColumn = "owner_id", entityColumn = "user_id")
    var user: User? = null
)

在哪里

owner_id 是 User 的 id定义于 Pet类(class)

user_id 与 User 中定义的用户 ID 匹配类(class)