基于自定义键字段名的外接联接

我想得到 join 使用异位素工作。exto似乎默认为'id'作为连接条件的键字段。尝试使用架构中的外键设置将其更改为列出用户ID和用户ID,但似乎不起作用。我错过了什么?

query =
          from(i in SgListItems,
            join: l in assoc(i, :sg_lists),
            join: u in assoc(i, :sg_users),
           select: [l.list_id, i.item_id]
          )

翻译成

SELECT s1."list_id", s0."item_id" FROM "sg_list_items" AS s0 INNER JOIN "sg_lists" AS s1 ON **s1."id"** = s0."list_id" INNER JOIN "sg_users" AS s2 ON **s2."id"** = s0."author_id" ) []

例如,sg_的模式如下:

 @primary_key {:list_id, :binary_id, autogenerate: true}
  schema "sg_lists" do
    field(:list_title, :string, null: false)
    field(:list_description, :map, null: false)

    has_many(:sg_list_items, MyApp.SgListItems, foreign_key: :list_id)
    timestamps()
  end

下面是sg_list_项的架构。

 @primary_key {:item_id, :binary_id, autogenerate: true}
  schema "sg_list_items" do
    belongs_to(:sg_lists, MyApp.SgLists, foreign_key: :list_id, type: :binary_id)
    field(:item_hierarchy, :map, null: false)
    field(:item_title, :string, null: false)
    field(:item_description, :map, null: false)
    belongs_to(:sg_users, MyApp.SgUsers, foreign_key: :author_id, type: :binary_id)
    timestamps()
  end