为我的模型获取持久的字段定义答案

【问题标题】：get persistent's field definitions for my model为我的模型获取持久的字段定义
【发布时间】：2018-03-19 02:52:13
【问题描述】：

给定这样的持久模型定义：

mkPersist sqlSettings [persistLowerCase|
Person
    name String
    age Int
    deriving Show
|]

我正在寻找一种方法来获取此模型的字段定义。 FieldDef 看起来是一个很有前途的数据类型，但是 persistent 不会生成 getPersonFields :: [FieldDef] 函数，那怎么办呢？

如果这是一个 XY 问题 - 这是背景：如果出现 json 解析错误，我想返回某种元数据。如果我尝试发布一个新人但 json 错误，我想回复如下内容：

{
    "result": "error",
    "code": 8,
    "message": "Could not parse payload as person",
    "fields": [
        {
            "name": "name",
            "type": "string"
        },
        {
            "name": "age",
            "type": "int"
        }
    ]
}

【问题讨论】：

标签： haskell yesod haskell-persistent

【解决方案1】：

一个展示如何完成的示例：

#!/usr/bin/env stack
{- stack
     --resolver lts-9.0
     --install-ghc
     runghc
     --package persistent
     --package persistent-sqlite
     --package persistent-template
-}

{-# LANGUAGE EmptyDataDecls             #-}
{-# LANGUAGE FlexibleContexts           #-}
{-# LANGUAGE GADTs                      #-}
{-# LANGUAGE GeneralizedNewtypeDeriving #-}
{-# LANGUAGE MultiParamTypeClasses      #-}
{-# LANGUAGE OverloadedStrings          #-}
{-# LANGUAGE QuasiQuotes                #-}
{-# LANGUAGE TemplateHaskell            #-}
{-# LANGUAGE TypeFamilies               #-}
import           Control.Monad.IO.Class  (liftIO)
import           Database.Persist
import           Database.Persist.Sqlite
import           Database.Persist.TH

share [mkPersist sqlSettings, mkMigrate "migrateAll"] [persistLowerCase|
Person
    name String
    age Int Maybe
    deriving Show
BlogPost
    title String
    authorId PersonId
    deriving Show
|]

main :: IO ()
main = let efields = entityFields $ entityDef (undefined :: Maybe Person)
       in print efields

演示：

$ stack efield.hs
[FieldDef {fieldHaskell = HaskellName {unHaskellName = "name"}, fieldDB = DBName {unDBName = "name"}, fieldType = FTTypeCon Nothing "String", fieldSqlType = SqlString, fieldAttrs = [], fieldStrict = True, fieldReference = NoReference},FieldDef {fieldHaskell = HaskellName {unHaskellName = "age"}, fieldDB = DBName {unDBName = "age"}, fieldType = FTTypeCon Nothing "Int", fieldSqlType = SqlInt64, fieldAttrs = ["Maybe"], fieldStrict = True, fieldReference = NoReference}]

【讨论】：