| Safe Haskell | None |
|---|---|
| Language | Haskell2010 |
Data.AppContainer.TH
Documentation
Arguments
| :: Options | Encoding options. |
| -> Name | Name of the type for which to generate |
| -> Q [Dec] |
Generates both ToJSON and FromJSON instance declarations for the given
data type.
This is a convienience function which is equivalent to calling both
deriveToJSON and deriveFromJSON.
Arguments
| :: Options | Encoding options. |
| -> Name | Name of the type for which to generate a |
| -> Q [Dec] |
Generates a ToJSON instance declaration for the given data type.
deriveJSONOptions :: String -> Options Source
class FromJSON a where
A type that can be converted from JSON, with the possibility of failure.
When writing an instance, use empty, mzero, or fail to make a
conversion fail, e.g. if an Object is missing a required key, or
the value is of the wrong type.
An example type and instance:
@{-# LANGUAGE OverloadedStrings #-}
data Coord = Coord { x :: Double, y :: Double }
instance FromJSON Coord where
parseJSON (Object v) = Coord <$>
v .: "x" <*>
v .: "y"
-- A non-Object value is of the wrong type, so use mzero to fail.
parseJSON _ = mzero
@
Note the use of the OverloadedStrings language extension which enables
Text values to be written as string literals.
Instead of manually writing your FromJSON instance, there are three options
to do it automatically:
- Data.Aeson.TH provides template-haskell functions which will derive an instance at compile-time. The generated instance is optimized for your type so will probably be more efficient than the following two options:
- Data.Aeson.Generic provides a generic
fromJSONfunction that parses to any type which is an instance ofData. - If your compiler has support for the
DeriveGenericandDefaultSignatureslanguage extensions,parseJSONwill have a default generic implementation.
To use this, simply add a deriving clause to your datatype and
declare a GenericFromJSON instance for your datatype without giving a definition
for parseJSON.
For example the previous example can be simplified to just:
@{-# LANGUAGE DeriveGeneric #-}
import GHC.Generics
data Coord = Coord { x :: Double, y :: Double } deriving Generic
instance FromJSON Coord @
Note that, instead of using DefaultSignatures, it's also possible
to parameterize the generic decoding using genericParseJSON applied
to your encoding/decoding Options:
instance FromJSON Coord where
parseJSON = genericParseJSON defaultOptions
Minimal complete definition
Nothing
Instances
class ToJSON a where
A type that can be converted to JSON.
An example type and instance:
@{-# LANGUAGE OverloadedStrings #-}
data Coord = Coord { x :: Double, y :: Double }
instance ToJSON Coord where
toJSON (Coord x y) = object ["x" .= x, "y" .= y]
@
Note the use of the OverloadedStrings language extension which enables
Text values to be written as string literals.
Instead of manually writing your ToJSON instance, there are three options
to do it automatically:
- Data.Aeson.TH provides template-haskell functions which will derive an instance at compile-time. The generated instance is optimized for your type so will probably be more efficient than the following two options:
- Data.Aeson.Generic provides a generic
toJSONfunction that accepts any type which is an instance ofData. - If your compiler has support for the
DeriveGenericandDefaultSignatureslanguage extensions (GHC 7.2 and newer),toJSONwill have a default generic implementation.
To use the latter option, simply add a deriving clause to your
datatype and declare a GenericToJSON instance for your datatype without giving a
definition for toJSON.
For example the previous example can be simplified to just:
@{-# LANGUAGE DeriveGeneric #-}
import GHC.Generics
data Coord = Coord { x :: Double, y :: Double } deriving Generic
instance ToJSON Coord @
Note that, instead of using DefaultSignatures, it's also possible
to parameterize the generic encoding using genericToJSON applied
to your encoding/decoding Options:
instance ToJSON Coord where
toJSON = genericToJSON defaultOptions
Minimal complete definition
Nothing
Instances