Elixir Access.filter用法及代碼示例

@spec filter((term() -> boolean())) ::
  access_fun(data :: list(), current_value :: list())

返回一個函數，該函數訪問列表中與提供的謂詞匹配的所有元素。

返回的函數通常作為訪問器傳遞給 Kernel.get_in/2 、 Kernel.get_and_update_in/3 和朋友。

例子

iex> list = [%{name: "john", salary: 10}, %{name: "francine", salary: 30}]
iex> get_in(list, [Access.filter(&(&1.salary > 20)), :name])
["francine"]
iex> get_and_update_in(list, [Access.filter(&(&1.salary <= 20)), :name], fn prev ->
...>   {prev, String.upcase(prev)}
...> end)
{["john"], [%{name: "JOHN", salary: 10}, %{name: "francine", salary: 30}]}

filter/1 也可用於從列表中彈出元素或列表中的鍵：

iex> list = [%{name: "john", salary: 10}, %{name: "francine", salary: 30}]
iex> pop_in(list, [Access.filter(&(&1.salary >= 20))])
{[%{name: "francine", salary: 30}], [%{name: "john", salary: 10}]}
iex> pop_in(list, [Access.filter(&(&1.salary >= 20)), :name])
{["francine"], [%{name: "john", salary: 10}, %{salary: 30}]}

如果沒有找到匹配項，則返回一個空列表，並且永遠不會調用更新函數

iex> list = [%{name: "john", salary: 10}, %{name: "francine", salary: 30}]
iex> get_in(list, [Access.filter(&(&1.salary >= 50)), :name])
[]
iex> get_and_update_in(list, [Access.filter(&(&1.salary >= 50)), :name], fn prev ->
...>   {prev, String.upcase(prev)}
...> end)
{[], [%{name: "john", salary: 10}, %{name: "francine", salary: 30}]}

如果謂詞不是函數或元數不正確，則會引發錯誤：

iex> get_in([], [Access.filter(5)])
** (FunctionClauseError) no function clause matching in Access.filter/1

如果訪問的結構不是列表，則會引發錯誤：

iex> get_in(%{}, [Access.filter(fn a -> a == 10 end)])
** (RuntimeError) Access.filter/1 expected a list, got: %{}