Files
graphql/src/Language/GraphQL.hs
Eugen Wissner 22313d05df Deprecate Language.GraphQL.Execute.Schema
It is not a schema (at least not a complete one), but a resolver list,
and the resolvers should be provided by the user separately, because the
schema can originate from a GraphQL document. Schema name should be free
to provide a data type for the real schema later.
2019-08-30 07:26:04 +02:00

36 lines
1.3 KiB
Haskell

-- | This module provides the functions to parse and execute @GraphQL@ queries.
module Language.GraphQL
( graphql
, graphqlSubs
) where
import Control.Monad.IO.Class (MonadIO)
import qualified Data.Aeson as Aeson
import Data.List.NonEmpty (NonEmpty)
import qualified Data.Text as T
import Language.GraphQL.Error
import Language.GraphQL.Execute
import Language.GraphQL.Parser
import qualified Language.GraphQL.Schema as Schema
import Text.Megaparsec (parse)
-- | If the text parses correctly as a @GraphQL@ query the query is
-- executed using the given 'Schema.Resolver's.
graphql :: MonadIO m
=> NonEmpty (Schema.Resolver m) -- ^ Resolvers.
-> T.Text -- ^ Text representing a @GraphQL@ request document.
-> m Aeson.Value -- ^ Response.
graphql = flip graphqlSubs $ const Nothing
-- | If the text parses correctly as a @GraphQL@ query the substitution is
-- applied to the query and the query is then executed using to the given
-- 'Schema.Resolver's.
graphqlSubs :: MonadIO m
=> NonEmpty (Schema.Resolver m) -- ^ Resolvers.
-> Schema.Subs -- ^ Variable substitution function.
-> T.Text -- ^ Text representing a @GraphQL@ request document.
-> m Aeson.Value -- ^ Response.
graphqlSubs schema f
= either parseError (execute schema f)
. parse document ""