Example of how to parse exiftool JSON output in Haskell

问题

I can't make sense of any of the documentation. Can someone please provide an example of how I can parse the following shortened exiftool output using the Haskell module Text.JSON? The data is generating using the command 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
}]

回答1:

Well, the easiest way is to get back a JSValue from the json package, like so (assuming your data is in text.json):

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))]})])

this just gives you a generic json Haskell data type.

The next step will be to define a custom Haskell data type for your data, and write an instance of JSON for that, that converts between JSValue's as above, and your type.

回答2:

Thanks to all. From your suggestions I was able to put together the following which translates the JSON back into name-value pairs.

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).

来源：https://stackoverflow.com/questions/2089194/example-of-how-to-parse-exiftool-json-output-in-haskell

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json

haskell

exiftool