Question

我正在尝试使用Yesod并持久创建一个网站。我对如何使用持久性API感到有点困惑。

这是我的两张表

Feed
    url Text
    UniqueFeed url

Subscription
    feed FeedId
    title Text
    UniqueSubscription feed

如果订阅源不存在，我正在尝试创建Feed，如果订阅尚不存在，则添加订阅该订阅源。

postFeedR :: Handler RepHtml
postFeedR = do
    url <- runInputPost $ ireq urlField "url"
    title <- runInputPost $ ireq textField "title"

    runDB $ do
        feedId <- insertFeed $ UniqueFeed url
        subscriptionId <- insertSubscription feedId title
        return 

    defaultLayout [whamlet| <p>done|]

insertFeed url = do
    f <- insertBy $ UniqueFeed url
    case f of
        Left (Entity uid _) -> uid
        Right (Key uid) -> do
            (Key uid) <- insert $ Feed url
            return uid

insertSubscription feedId title = do
    s <- insertBy $ UniqueSubscription feedId
    case s of 
        Left (Entity uid _) -> uid
        Right (Key uid) -> do
            (Key uid) <- insert $ Subscription feedId title
            return uid

我收到以下错误。我不明白为什么ghc认为insertFeed和insertSubscription的返回值应该是UniqueFeed和UniqueSubscription。我希望这些函数返回新创建的记录的键。

此外，我似乎丢掉了我在案件的每个右边条款中得到的密钥。为什么持久性返回那些键。如果UniqueSubscription不在数据库中，则持久性没有足够的信息来创建新的订阅记录，因为它缺少标题，而不在UniqueSubscription上。

如果有人能给我一些关于如何使用持久性API的指示，我真的很感激。

Handler/Home.hs:62:9:
    Kind incompatibility when matching types:
      a0 :: *
      GHandler App App :: * -> *
    Expected type: (a0 -> t0)
                   -> (t0 -> a0 -> m0 a0) -> YesodDB App App (m0 a0)
      Actual type: (a0 -> t0) -> (t0 -> a0 -> m0 a0) -> a0 -> m0 a0
    In a stmt of a 'do' block: feedId <- insertFeed $ UniqueFeed url
    In the second argument of `($)', namely
      `do { feedId <- insertFeed $ UniqueFeed url;
            subscriptionId <- insertSubscription feedId title;
            return }'

Handler/Home.hs:62:9:
    Couldn't match type `YesodPersistBackend App' with `(->)'
    Expected type: (a0 -> t0)
                   -> (t0 -> a0 -> m0 a0) -> YesodDB App App (m0 a0)
      Actual type: (a0 -> t0) -> (t0 -> a0 -> m0 a0) -> a0 -> m0 a0
    In a stmt of a 'do' block: feedId <- insertFeed $ UniqueFeed url
    In the second argument of `($)', namely
      `do { feedId <- insertFeed $ UniqueFeed url;
            subscriptionId <- insertSubscription feedId title;
            return }'

Handler/Home.hs:74:20:
    Couldn't match expected type `Unique Feed'
                with actual type `Database.Persist.Store.PersistValue'
    In the first argument of `return', namely `uid'
    In a stmt of a 'do' block: return uid
    In the expression:
      do { (Key uid) <- insert $ Feed url;
           return uid }

Handler/Home.hs:83:20:
    Couldn't match expected type `Unique Subscription'
                with actual type `Database.Persist.Store.PersistValue'
    In the first argument of `return', namely `uid'
    In a stmt of a 'do' block: return uid
    In the expression:
      do { (Key uid) <- insert $ Subscription feedId title;
           return uid }

Answer 1

insertBy不会将唯一约束作为参数，getBy更合适。

但是 ~~insertUnique~~ 与Maybe结果的可能性很小。

postFeedR :: Handler RepHtml
postFeedR = do
    url <- runInputPost $ ireq urlField "url"
    title <- runInputPost $ ireq textField "title"

    runDB $ do
        feedId <- insertFeed url
        _mbSubscriptionId <- insertUnique $ Subscription feedId title
        return ()

    defaultLayout ...

insertFeed url = do
    f <- insertBy $ Feed url
    case f of
        Left (Entity uid _) -> return uid
        Right uid -> return uid

Haskell Persistent插入行（如果尚未在数据库中）

1 个答案: