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Example of how to parse exiftool JSON output in Haskell

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
}]
like image 495
me2 Avatar asked Jan 18 '10 21:01

me2


2 Answers

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.

like image 61
Don Stewart Avatar answered Oct 17 '22 02:10

Don Stewart


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).

like image 24
me2 Avatar answered Oct 17 '22 03:10

me2