Haskell 不能'；t匹配预期类型‘；邮政’；实际类型为‘；路线应用程序&x2019；

我正在尝试使用函数

postBody

postBody :: (Yesod site, RedirectUrl site url) => url -> ByteString -> YesodExample site () 

从包裹里拿出来。文件上说它可以这样使用

import Data.Aeson
postBody HomeR (encode $ object ["age" .= (1 :: Integer)])

但是当我尝试在我的应用程序中使用它时

    describe "Posts" $ do
        it "posts post and returns post" $ do
            postBody PostR (encode $ object [
                "body" .= ("test post" :: Text),
                "title" .= ("test post" :: Text),
                "coverImage" .= ("test post" :: Text),
                "author" .= (0 :: Int)
                ])
            statusIs 200

我弄错了

• Couldn't match expected type ‘Post’ with actual type ‘Route App’
• In the first argument of ‘postBody’, namely ‘PostR’
  In a stmt of a 'do' block:
    postBody
      PostR
      (encode
       $ object
           ["body" .= ("test post" :: Text), "title" .= ("test post" :: Text),

            "coverImage" .= ("test post" :: Text), "author" .= (0 :: Int)])
  In the second argument of ‘($)’, namely
    ‘do { postBody
            PostR
            (encode
             $ object
                 ["body" .= ("test post" :: Text), "title" .= ("test post" :: Text),
                  ....]);
          statusIs 200 }’

我的用法似乎与示例相同，所以我看不出它为什么会失败

PostR在这里的routes文件中

/PostR POST

以及它的处理程序

postPostR :: Handler Value
postPostR = do
    post <- requireJsonBody :: Handler Post
    maybeUserID <- maybeAuthId
    case maybeUserID of
        Just userID -> do
            let post' = post {postAuthor = userID}
            inserted <- runDB $ insertEntity post'
            returnJson inserted
        Nothing -> sendResponseStatus status403 ("Unauthorized" :: Text)

pospostost:：处理程序值
postostr=do
post sendResponseStatus 403（“未授权”：：文本）

问题在于posts表中有一个列主体，因此Yesod为此创建了postBody，这与Yesod.Test中的postBody冲突

解决方案是使用合格的yesod测试导入

import qualified Yesod.Test as T

---

您的

PostR

是如何定义的？@FyodorSoikin被编辑以显示信息。