无法将类型"PersistEntityBackend(实体a)"与"SqlBackend"进行匹配 [英] Couldn't match type ‘PersistEntityBackend (Entity a)’ with ‘SqlBackend’

问题描述

我有以下内容:

asSqlBackendReader :: ReaderT SqlBackend m a -> ReaderT SqlBackend m a
asSqlBackendReader = id

insertEnt :: (Entity a) -> IO (Key (Entity a))
insertEnt x = runWithDb $ do
  insert $ x
  where runWithDb = runSqlite "test.db" . asSqlBackendReader

asSqlBAckendReader 的目的是由于.

我遇到了以下错误:

• Couldn't match type ‘PersistEntityBackend (Entity a)’
                 with ‘SqlBackend’
    arising from a use of ‘insert’
• In a stmt of a 'do' block: insert $ x
  In the second argument of ‘($)’, namely ‘do { insert $ x }’
  In the expression: runWithDb $ do { insert $ x }

推荐答案

将约束添加到 insertEnt 的签名中.您还需要一个 PersistEntity 约束.

Add the constraint to the signature of insertEnt. You'll also need a PersistEntity constraint.

insertEnt
  :: ( PersistEntity (Entity a)
     , PersistEntityBackend (Entity a) ~ SqlBackend)
  => Entity a -> IO (Key (Entity a))

---

要推断出这一点(不只是向编译器提供其间接要求的内容)，您可以查看我们也有

type PersistRecordBackend record backend =
  ( PersistEntity record
  , PersistEntityBackend record ~ BaseBackend backend)

此外，在您的应用程序中，您有一些具体类型:

Furthermore, in your application you have some concrete types:

backend ~ SqlBackend
m ~ (some concrete transformer stack on top of IO)
record ~ Entity a

这些具体类型释放 PersistStoreWrite后端和 MonadIO m ，并且由于 Entity a 仍然大部分都是抽象的，因此您将面临两个约束定义 PersistRecordBackend .实际上，您可以使用该同义词作为速记:

These concrete types discharge PersistStoreWrite backend and MonadIO m, and since Entity a is still mostly abstract, you are left with the two constraints that define PersistRecordBackend. You could in fact use that synonym as a shorthand:

insertEnt
  :: PersistRecordBackend (Entity a) SqlBackend
  => Entity a -> IO (Key (Entity a))

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