如何获得具有超载字段名称的经典镜头？

我正在尝试为具有相同字段名称的记录构建镜头。除此之外，我试图“包装/扩展”这些基本记录，并希望相同的字段名称适用于包装/扩展的记录（我相信，优雅的镜头就是这样做的）。我如何让以下内容发挥作用：

-- Data types for context of the code snippet below

data Download = Download {
  userId :: UserId
  ,gid :: Gid
  ,logId :: LogId
  ,parentId :: Maybe DownloadId
  ,createdAt :: UTCTime
  ,updatedAt :: UTCTime
}

data File = File {
  downloadId :: DownloadId
  ,fpath :: String
  ,len :: Int
  ,createdAt :: UTCTime
  ,updatedAt :: UTCTime
}

data Url = Url {
  downloadId :: DownloadId
  ,fileId :: FileId
  ,url :: URL
  ,createdAt :: UTCTime
  ,updatedAt :: UTCTime
}

data DownloadObject = DownloadObject {
  _key :: DownloadId
  ,_dbDownload :: Download
  ,_dbFiles :: [FileObjects]
}

data FileObject = FileObject {
  _key :: FileId
  ,_dbFile :: File,
  ,_dbUrls :: [UrlObjects]
}

data UrlObject = UrlObject {
  _key :: UrlId
  ,_dbUrl :: Url
}

fetchDownload :: DownloadId -> DownloadObject

考虑到这些数据类型，我如何让以下镜头发挥作用：

dload <- fetchDownload dloadId
dload ^. key -- of type DownloadId
dload ^. createdAt -- of type UTCTime
((dload ^. files) !! 1) ^. key -- of type FileId 
((dload ^. files) !! 1) ^. createdAt -- of type UTCTime

Use makeFields from Control.Lens.TH https://hackage.haskell.org/package/lens-4.14/docs/Control-Lens-TH.html:

{-# language
  TemplateHaskell, MultiParamTypeClasses, FunctionalDependencies, FlexibleInstances #-}

import Control.Lens

data Bar a n = Bar {
  _barAge :: a,
  _barName :: n
} deriving Show

data Foo = Foo {
  _fooAge  :: Int,
  _fooName :: String
} deriving Show

makeFields ''Bar
makeFields ''Foo

现在您有以下课程：

class HasName s a | s -> a where
  name :: Lens' s a

class HasAge s a | s -> a where
  age :: Lens' s a

以及适当的实例Foo and Bar。例子：

> Foo 10 "foo" ^. age
10
> Bar 10 "bar" ^. age
10

然后，您可以实现包装对象的类。例子：

data FooWrapper = FooWrapper {
  _fooWrapperKey :: Int,
  _fooWrapperFoo :: Foo
} deriving Show

makeFields ''FooWrapper

instance HasName FooWrapper String where
  name = foo . name

instance HasAge FooWrapper Int where
  age = foo . age