消除歧义哈斯克尔埃森的接线员

我试图在将对象序列化为json时重命名键。

我知道这样做的方法是,而不仅仅是使用派生泛型,用自定义键名定义一个实例,如下所示:

-- instance ToJSON R2  -- old
instance ToJSON R2 where
  toJSON (R2 recCode recDate) = object [ "code" .= recCode , "date" .= recDate ]
-- new

然而,这给了我:

<interactive>:2:70: error:
    Ambiguous occurrence ‘.=’
    It could refer to either ‘Data.Aeson..=’, imported from ‘Data.Aeson’ (and originally defined in ‘aeson-1.3.1.1:Data.Aeson.Types.ToJSON’)
                          or ‘Control.Lens..=’, imported from ‘Control.Lens’ (and originally defined in ‘Control.Lens.Setter’)

我试图解决这个问题,是为了明确地迫使 .= 运算符,例如:

(.=) = Data.Aeson.(.=)

这是一个猜测,但似乎是错误的语法。我通过类比将parens添加到以下资源中:

• https://hackage.haskell.org/package/aeson-0.6.1.0/docs/src/Data-Aeson-Types-Class.html#.%3D
• http://www.stephendiehl.com/posts/protolude.html
• https://stackoverflow.com/a/18372384/1052117

这给了我一个错误:

(.=) = Data.Aeson (.=)

<interactive>:1:8: error:
    Not in scope: data constructor ‘Data.Aeson’
    No module named ‘Data’ is imported.

“let”的正确语法是什么 .= 明确地成为 .= 从…起 Data.Aeson " ?