Haskell Servant将自定义数据传递给auth处理程序

Rey*_*och 4 database authentication haskell pool servant

我正在使用自定义monad(带阅读器)轻松地将数据库(如DB池)传递给我的处理程序(在使用自定义monad之前,我曾将连接作为fn参数传递).

这就是我定义自定义monad的方法:

newtype Controller a = Controller
    { runController :: ReaderT ServerEnvironment Handler a
    } deriving ( Functor, Applicative, Monad, MonadReader ServerEnvironment, 
                 MonadError ServantErr, MonadIO )
Run Code Online (Sandbox Code Playgroud)

ServerEnvironment只是我用来携带数据的自定义数据类型.

问题是,对于我,AuthHandler我必须具体使用以下功能:

r -> Handler usr
Run Code Online (Sandbox Code Playgroud)

作为身份验证处理程序,我不能使用我的自定义处理程序,它将是:

r -> Controller usr
Run Code Online (Sandbox Code Playgroud)

而且我也无法传递我,ConnectionPool因为签名不能是:

ConnPool -> r -> Handler usr
Run Code Online (Sandbox Code Playgroud)

那么,如何在不使用全局IO状态的情况下将额外数据传递给servant中的身份验证处理程序?

use*_*650 7

AuthHandler你放入上下文没有在顶层进行定义!通常,您需要这样做,main以便您可以访问您创建的数据库连接等:

type API = 
  ... :<|> (AuthProtect "myProtection" :> ...) :<|> ...

type instance AuthServerData (AuthProtect "myProtection") = User

server :: ServerEnvironment -> Server API
server env = ...

setupEnv :: IO ServerEnvironment
setupEnv = ..

-- This is essentially a 'Controller'.
authenticate :: ServerEnvironment -> Handler User
authenticate conn = ...

main :: IO ()
main = do
  env <- setupEnv
  -- Now, because we have access to the env, we can turn our
  -- 'authenticate' into the right type before putting it
  -- in the context
  let ctx = authenticate env :. EmptyContext
  run 8080 $ serveWithContext myAPI (server conn) ctx
Run Code Online (Sandbox Code Playgroud)