XML to JSON with Scala

Question

For a XML snippet like this:

val fruits =
<fruits>
  <fruit>
    <name>apple</name>
    <taste>ok</taste>
  </fruit>
  <fruit>
    <name>banana</name>
    <taste>better</taste>
  </fruit>
</fruits>

doing something like:

fruits \\ "fruit"

will return a sequence of type scala.xml.NodeSeq with all the fruits and sub nodes inside.

What is the best way to convert this to a list of JSON objects? I'm trying to send my list of fruits back to a browser. I had a look at scala.util.parsing.json.JSONObject and scala.util.parsing.json.JSONArray, but I don't know how to get from NodeSeq to anyone of the latter.

If at all possible, I would love to see how it's done with plain Scala code.

Community · Accepted Answer · 2017-05-23 12:00:06Z

17

This might be relevant. Here is my solution using spray-json:

import scala.xml._
import cc.spray.json._
import cc.spray.json.DefaultJsonProtocol._

implicit object NodeFormat extends JsonFormat[Node] {
  def write(node: Node) =
    if (node.child.count(_.isInstanceOf[Text]) == 1)
      JsString(node.text)
    else
      JsObject(node.child.collect {
        case e: Elem => e.label -> write(e)
      }: _*)

  def read(jsValue: JsValue) = null // not implemented
}

val fruits =
  <fruits>
    <fruit>
      <name>apple</name>
      <taste>
        <sweet>true</sweet>
        <juicy>true</juicy>
      </taste>
    </fruit>
    <fruit>
      <name>banana</name>
      <taste>better</taste>
    </fruit>
  </fruits>

val json = """[{"name":"apple","taste":{"sweet":"true","juicy":"true"}},{"name":"banana","taste":"better"}]"""

assert((fruits \\ "fruit").toSeq.toJson.toString == json)

edited May 23, 2017 at 12:00

CommunityBot

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answered Mar 1, 2012 at 21:31

elbowich

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7 Comments

Jack Over a year ago

This looks great, but your sample does not run. I get the following: error: not found: value JsField.

elbowich Over a year ago

@JacobusR Oops, still using 1.0.1 version of spray-json. There is indeed no JsField class in 1.1.0. Replaced it with a tuple.

Jack Over a year ago

Works like a charm, thanks. Always lovely to get a runnable bit of code, rather than just a snippet (+1 for that :-).

crockpotveggies Over a year ago

FYI SprayJson has been recently updated, looks like JsonFormat has moved

fommil Over a year ago

A JsonWriter is more appropriate here as it means you don't need to have a dangling unimplemented read.

|

Thomas Luechtefeld · Accepted Answer · 2018-07-02 23:49:48Z

3

org.json4s makes this pretty simple:

import org.json4s.Xml.toJson

val fruits =
  <fruits>
    <fruit>
      <name>apple</name>
      <taste>
        <sweet>true</sweet>
        <juicy>true</juicy>
      </taste>
    </fruit>
    <fruit>
      <name>banana</name>
      <taste>better</taste>
    </fruit>
  </fruits>

println(toJson(fruits))

answered Jul 2, 2018 at 23:49

Thomas Luechtefeld

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Comments

om-nom-nom · Accepted Answer · 2012-03-29 16:29:49Z

2

I think you should use ScalaXB to turn the XML into scala classes and then write the appropriate toJson bits to output Json.

Should work a treat.

edited Mar 29, 2012 at 16:29

om-nom-nom

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answered Mar 1, 2012 at 16:38

tylerweir

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1 Comment

om-nom-nom Over a year ago

The problem with ScalaXB is that it requires XML schema

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