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Elm: Json decoder timestamp to Date

Tags:

elm

I'm trying to convert a timestamp (ex: "1493287973015") from a JSON to a Date type.

So far I created this custom decoder:

stringToDate : Decoder String -> Decoder Date
stringToDate decoder =
  customDecoder decoder Date.fromTime

But it doesn't work because it has return a Result, not a Date:

Function `customDecoder` is expecting the 2nd argument to be:

    Time.Time -> Result String a

But it is:

    Time.Time -> Date.Date

Is there a way to do a conversion?

like image 797
Guilhem Soulas Avatar asked May 10 '16 18:05

Guilhem Soulas


2 Answers

Assuming your JSON is actually placing the numeric value inside quotes (meaning you are parsing the JSON value "1493287973015" and not 1493287973015), your decoder could look like this:

import Json.Decode exposing (..)
import Date
import String

stringToDate : Decoder Date.Date
stringToDate =
  string
    |> andThen (\val ->
        case String.toFloat val of
          Err err -> fail err
          Ok ms -> succeed <| Date.fromTime ms)

Notice that stringToDate doesn't get passed any parameters, as opposed to your example in which you were attempting to pass a Decoder String as a parameter. That's not quite how decoders work.

Instead, this can be done by building upon more primitive decoders, in this case, we start with the decoder string from Json.Decode.

The andThen portion then takes the string value given by the decoder, and tries to parse it to a float. If it is a valid Float, it is fed into Date.fromTime, otherwise, it's a failure.

The fail and succeed functions wrap up the normal values you're dealing with into Decoder Date.Date contexts so they can be returned.

like image 183
Chad Gilbert Avatar answered Nov 06 '22 13:11

Chad Gilbert


Two things, a JSON may actually have the milliseconds as an integer, not a string and things have changed since v 0.19 of Elm.

Given that your JSON looks something like.

{
    ...
    "someTime": 1552483995395,
    ...
}

Then this would decode to a Time.Posix:

import Json.Decode as Decode

decodeTime : Decode.Decoder Time.Posix
decodeTime =
    Decode.int
        |> Decode.andThen
            (\ms ->
                Decode.succeed <| Time.millisToPosix ms
            )
like image 33
PerLundholm Avatar answered Nov 06 '22 12:11

PerLundholm