Question

我以这种方式在我的项目中使用Amnesia：

use Amnesia

defdatabase Database do
  deftable Account, [{:id, autoincrement}, :email, :account_number, :password, :amount],
    type: :set, index: [:email, :account_number] do
      @type t :: %Account{
        id: non_neg_integer,
        email: String.t,
        account_number: String.t,
        amount: integer
      }

      def money_transfer(from_account_number, to_account_number, amount) do
        from_account = Account.where(account_number == from_account_number)
        ...
      end
  end
end

在我的函数money_transfer/3中，我需要在Account表上获取帐户，但函数where不起作用：

warning: variable "account_number" does not exist and is being expanded to "account_number()", please use parentheses to remove the ambiguity or change the variable name
  lib/Database.ex:22


== Compilation error in file lib/Database.ex ==
** (CompileError) lib/Database.ex:22: undefined function account_number/0
    (stdlib) lists.erl:1338: :lists.foreach/2
    lib/Database.ex:4: (module)
➜

如何解决？我不知道为什么会出现此错误，因为account_number是deftable上声明的列。我跟着Amnesia readme。

Answer 1

Reading the tests of Amnesia，我找到了解决方案：

defdatabase Database do
  deftable Account, [{:id, autoincrement}, :email, :account_number, :password, :amount],
    type: :set, index: [:email, :account_number] do

    ...

    def get_account_by_account_number(by_account_number) do
      accounts_by_account_number = where(account_number == by_account_number)

      case Amnesia.Selection.values(accounts_by_account_number) do
        [account] -> {:ok, account}
        [] -> {:error, "not found account"}
      end
    end

当你在detable内创建一个函数时，你不应该使用它表的名称。您应该只使用where函数。

我不太清楚为什么会这样做。

如何使用Amnesia（又名精致选择器）？

1 个答案: