Nested maps with different structure

Question

I am new to Scala and I want to build an object that looks like below and then convert to Json, but doing something like this gives me error. How can I achieve this?

This is what I tried:

Map(a -> Map(b-> "1", c -> "2"),
Map(d -> values.map(v => Map(b-> v.value1, c -> v.value2)))
.asJson

Expected outcome:

{
"a":{"b": "1", "c": "2"},
"d":[{"b":"x1","c":"x2"},{"b":x3,"c":"x4"}...]
}

In general, you would want to model your data using case class and then convert that to a Json, or if this is something used in a single place then you may indeed consider using the Json type provided by your library rather than using an heterogeneous map. — Luis Miguel Mejía Suárez
– Luis Miguel Mejía Suárez, Commented Oct 26, 2022 at 13:00
It is used in just one place. What do you mean by "your library" @LuisMiguelMejíaSuárez? can you provide any example? — user2720919
– user2720919, Commented Oct 26, 2022 at 17:59
You are using some kind of library to manage JSON, be it circe, uPickle, zio-json, etc. Each one of those would have a proper way to build a Json object, just like the accepted answer shows how to do it using circe. — Luis Miguel Mejía Suárez
– Luis Miguel Mejía Suárez, Commented Oct 26, 2022 at 18:42

Alex · Accepted Answer · 2022-10-26 04:30:18Z

2

You should reserve to using the JSON library's provided DSL instead, i.e. circe JSON DSL:

import io.circe.Json
import io.circe.syntax.*

Json.obj(
  "a" := Json.obj("b" := "1", "c" := "2")
  "d" := values.map(v => Json.obj("b" := v.value1, "c" := v.value2))
)

answered Oct 26, 2022 at 4:30

Alex

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