在 where 语句中使用声明引号

Question

我正在实施基于使用标准 haskell functions/combinators 构建数据库查询的 DSL。从一个实现 POV 我决定像这样在查询中表示变量：

newtype Variable = Var { fromVar :: Text }

然而，这迫使用户经常写 Var "something"，所以我决定编写一个自动执行此操作的准引号。

这是 DSL 的示例：

{-# LANGUAGE OverloadedStrings #-}
maxQuery :: Query MAX
maxQuery = match
         ( sch `isa` "school"
         $ forWhich "ranking" `labelMatches` ran $ε)
         `get` [ran]
         `max` [ran]
    where 
        [sch,ran] = map Var ["sch","ran"]

我希望它是什么：

maxQuery :: Query MAX
maxQuery = match
         ( sch `isa` "school"
         $ forWhich "ranking" `labelMatches` ran $ε)
         `get` [ran]
         `max` [ran]
    where [defVars| sch ran |]

或类似的内容。

我写的准引用在这里：

{-# LANGUAGE TemplateHaskell #-}
module TypeDBTH where 
import Language.Haskell.TH.Syntax
import Language.Haskell.TH.Quote
import Data.List.Split
import Data.Text (pack)

mkVars :: [String] -> Dec
mkVars vars = ValD 
                (ListP (map (VarP . mkName) vars))
                (NormalB (ListE (map (\v -> AppE (ConE $ mkName "Var")
                                          $ AppE (VarE $ mkName "pack")
                                                 (LitE $ StringL v))
                                     vars)))
                []

defVars :: QuasiQuoter
defVars = QuasiQuoter { quoteDec = quoteVars }
                 --, quoteExp = expQuoteVars }

quoteVars :: String -> Q [Dec]
quoteVars = return . return . mkVars . filter (/= "") . splitOn " "

expQuoteVars :: String -> Q Exp
expQuoteVars s = return $ LetE [(mkVars . filter (/= "") . splitOn " " $ s)] (LitE $ StringL "x")

本来我只写了quoteVars。为了在 ghci 中进行测试，我添加了 expQuoteVars。但是，现在删除后一个并尝试编写

...
    where [defVars| sch ran |]

给我留下了两个错误：

lib/TypeDBQuery.hs:806:1: error:
    parse error (possibly incorrect indentation or mismatched brackets)

因为 where [quasiquoter] 后面没有任何内容

和

lib/TypeDBQuery.hs:807:5: error:
    • Exception when trying to run compile-time code:
        lib/TypeDBTH.hs:18:11-46: Missing field in record construction quoteExp

      Code: Language.Haskell.TH.Quote.quoteExp defVars " sch ran "
    • In the quasi-quotation: [defVars| sch ran |]
    |
807 | x = [defVars| sch ran |] 
    |     ^^^^^^^^^^^^^^^^^^^^

如何使用准引号代替 quoteDec 而不是 quoteExp？这可能吗？

如果这样更容易的话，我也愿意像这样使用它：

maxQuery :: Query MAX
maxQuery = let [defVars | sch ran |] in
                $ match
                ( sch `isa` "school"
                $ forWhich "ranking" `labelMatches` ran $ε)
                `get` [ran]
                `max` [ran]

我查看了 wiki.haskell.org 和 TH 模块的“教程”和信息站点，但不知道如何执行此操作... https://wiki.haskell.org/Template_Haskell#What_to_do_when_you_can.27t_splice_that_there https://wiki.haskell.org/Quasiquotation https://wiki.haskell.org/A_practical_Template_Haskell_Tutorial

Answer 1

很遗憾，您只能在顶级声明中使用声明类引号。来自 the documentation:

A quasiquote may appear in place of

An expression

A pattern

A type

A top-level declaration

您可以考虑使用 OverloadedStrings:

而不是使用 TH

instance IsString Variable where
  fromString str = Var (pack str)

maxQuery :: Query MAX
maxQuery = match
         ( "sch" `isa` "school"
         $ forWhich "ranking" `labelMatches` "ran" $ε)
         `get` ["ran"]
         `max` ["ran"]

在 where 语句中使用声明引号

Using a declaration quoter in a where statement

haskell

template-haskell

quasiquotes