Get data from json in Scala

Question

I am trying to replicate this Python code in Scala:

import json

def test_json():
    parsed = json.loads("""[{"UserName":"user1","Tags":"one, two, three"},{"UserName":"user2","Tags":"one, two, three"}]""")
    tags = parsed[0]["Tags"].split(", ")
    print(tags)

test_json()

And I am coming up with this gibberish which does not work:

import scala.util.parsing.json._

def testSix: Seq[String] = {
    val parsed = JSON.parseFull("""[{"UserName":"user1","Tags":"one, two, three"},{"UserName":"user2","Tags":"one, two, three"}]""")
    parsed.map(_ match {
        case head :: tail => head
        case _ => Option.empty
    }).map(_ match {
        case m: Map[String, String] => m("Tags").split(", ")
        case _ => Option.empty
    })
}

How can I get the Tags in the first entry of the json?

Which library are you using to parse the JSON?

Yuval Itzchakov
– Yuval Itzchakov

2017-06-14 08:09:12 +00:00
Commented Jun 14, 2017 at 8:09 — Yuval Itzchakov
– Yuval Itzchakov, Commented Jun 14, 2017 at 8:09
@YuvalItzchakov updated.

blueFast
– blueFast

2017-06-14 08:10:15 +00:00
Commented Jun 14, 2017 at 8:10 — blueFast
– blueFast, Commented Jun 14, 2017 at 8:10

chengpohi · Accepted Answer · 2017-06-14 08:35:57Z

1

It's a not good way to use the raw scala JSON util to parse JSON, it's not type safe. maybe you can JSON4s or other library.

And there is a way to achieve this by using scala util:

  def testSix: Seq[String] = {
    val parsed = JSON.parseFull("""[{"UserName":"user1","Tags":"one, two, three"},{"UserName":"user2","Tags":"one, two, three"}]""")
    val typedResult: Option[List[Map[String, String]]] = parsed.map {
      case a: List[_] => {
        a.map {
          case t: Map[String, String] => t
        }
      }
    }
    typedResult.map(_.flatMap(_.get("Tags")).toSeq).getOrElse(Seq())
  }
  testSix.head

Explanation:

parse json firstly
try to convert the parsed result(type: Option[Any]) to type: Option[List[Map[String, String]]]

edited Jun 14, 2017 at 8:35

answered Jun 14, 2017 at 8:29

chengpohi

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2 Comments

blueFast Over a year ago

Thanks, close, returns List(one, two, three, one, two, three), but I need List(one, two, three)

chengpohi Over a year ago

testSix.head.split(",\\s+").toList

prayagupadhyay · Accepted Answer · 2017-06-14 09:28:17Z

0

First of all default scala.util.parsing.json._ library might be ugly in this case, still try something as below

scala> val terriblyTraversedProperty = parsed.map(_ match { case  head :: tail => head case _ => Option.empty }).map(_ match { case firstUser: Map[String, String] => firstUser("Tags") case _ => Option.empty })
terriblyTraversedProperty: Option[java.io.Serializable] = Some(one, two, three)

To split the property which is Option[String],

scala> val tags = terriblyTraversedProperty.map(_ match { case tagString: String => tagString.split(",").toList case _ => Option.empty[List[String]]} ).get
tags: Product with java.io.Serializable = List(one, " two", " three")

tags is Product with java.io.Serializable but is also an instance of List[String]

scala> tags.isInstanceOf[Seq[String]]
res35: Boolean = true

I can recommend scala json4s library if you have freedom to choose libraries,

add json4s to build.sbt,

libraryDependencies += "org.json4s" % "json4s-native_2.11" % "3.5.2"

example,

scala> import org.json4s._
import org.json4s._

scala> import org.json4s.native.JsonMethods._
import org.json4s.native.JsonMethods._

scala> val parsed = parse("""[{"UserName":"user1","Tags":"one, two, three"},{"UserName":"user2","Tags":"one, two, three"}]""")
parsed: org.json4s.JValue = JArray(List(JObject(List((UserName,JString(user1)), (Tags,JString(one, two, three)))), JObject(List((UserName,JString(user2)), (Tags,JString(one, two, three))))))

to get the property from first element

scala> parsed.values.asInstanceOf[List[Map[String, String]]](0)("Tags")
res13: String = one, two, three

edited Jun 14, 2017 at 9:28

answered Jun 14, 2017 at 8:31

prayagupadhyay

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5 Comments

blueFast Over a year ago

Thanks. The Some in Some(one, two, three) is what is killing me. I want Seq[String]. I can not change libraries unfortunately.

prayagupadhyay Over a year ago

Ok, whenever you see Some, do .map :)

prayagupadhyay Over a year ago

@delavnog See the answer, you will love it :)

blueFast Over a year ago

Better, thanks, but this is not Seq[String], but java.io.Serializable. I really need a simple Seq[String]

Dima Over a year ago

Your json has "one two three" - that's a string. If you want seq, change that to ["one", "two", "three"].

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