6

我无法理解任何文档。有人可以提供一个示例，说明如何exiftool使用 Haskell 模块解析以下缩短的输出Text.JSON吗？使用命令生成数据exiftool -G -j <files.jpg>。

[{
  "SourceFile": "DSC00690.JPG",
  "ExifTool:ExifToolVersion": 7.82,
  "File:FileName": "DSC00690.JPG",
  "Composite:LightValue": 11.6
},
{
  "SourceFile": "DSC00693.JPG",
  "ExifTool:ExifToolVersion": 7.82,
  "File:FileName": "DSC00693.JPG",
  "EXIF:Compression": "JPEG (old-style)",
  "EXIF:ThumbnailLength": 4817,
  "Composite:LightValue": 13.0
},
{
  "SourceFile": "DSC00694.JPG",
  "ExifTool:ExifToolVersion": 7.82,
  "File:FileName": "DSC00694.JPG",
  "Composite:LightValue": 3.7
}]
4

2 回答 2

10

好吧，最简单的方法是从json包中取回一个 JSValue ，如下所示（假设您的数据在 text.json 中）：

Prelude Text.JSON> s <- readFile "test.json"
Prelude Text.JSON> decode s :: Result JSValue
Ok (JSArray [JSObject (JSONObject {fromJSObject = [("SourceFile",JSString (JSONString {fromJSString = "DSC00690.JPG"})),("ExifTool:ExifToolVersion",JSRational False (391 % 50)),("File:FileName",JSString (JSONString {fromJSString = "DSC00690.JPG"})),("Composite:LightValue",JSRational False (58 % 5))]}),JSObject (JSONObject {fromJSObject = [("SourceFile",JSString (JSONString {fromJSString = "DSC00693.JPG"})),("ExifTool:ExifToolVersion",JSRational False (391 % 50)),("File:FileName",JSString (JSONString {fromJSString = "DSC00693.JPG"})),("EXIF:Compression",JSString (JSONString {fromJSString = "JPEG (old-style)"})),("EXIF:ThumbnailLength",JSRational False (4817 % 1)),("Composite:LightValue",JSRational False (13 % 1))]}),JSObject (JSONObject {fromJSObject = [("SourceFile",JSString (JSONString {fromJSString = "DSC00694.JPG"})),("ExifTool:ExifToolVersion",JSRational False (391 % 50)),("File:FileName",JSString (JSONString {fromJSString = "DSC00694.JPG"})),("Composite:LightValue",JSRational False (37 % 10))]})])

这只是为您提供了一个通用的 json Haskell 数据类型。

下一步是为你的数据定义一个自定义的 Haskell 数据类型，并为此编写一个 JSON 实例，在上面的 JSValue 和你的类型之间进行转换。

于 2010-01-18T21:35:44.110 回答
2

谢谢大家。根据您的建议，我能够汇总以下内容，将 JSON 转换回名称-值对。

data Exif = 
    Exif [(String, String)]
    deriving (Eq, Ord, Show)

instance JSON Exif where
    showJSON (Exif xs) = showJSONs xs
    readJSON (JSObject obj) = Ok $ Exif [(n, s v) | (n, JSString v) <- o]
        where 
            o = fromJSObject obj
            s = fromJSString

Unfortunately, it seems the library is unable to translate the JSON straight back into a simple Haskell data structure. In Python, it is a one-liner: json.loads(s).

于 2010-01-19T00:18:19.230 回答