Marvin Marvin - 1 year ago 145
JSON Question

How to use jackson to deserialize to Kotlin collections

Sample code what I want:

data class D(val a: String, val b: Int)
val jsonStr = """[{"a": "value1", "b": 1}, {"a": "value2", "b":"2}]"""
// what I need
val listOfD: List<D> = jacksonObjectMapper().whatMethodAndParameter?

Answer Source

With Jackson Kotlin Module current versions, if you import the full module package or the specific extension function you'll have all extension methods available. Such as:

import com.fasterxml.jackson.module.kotlin.*  
val JSON = jacksonObjectMapper()  // keep around and re-use
val myList: List<String> = JSON.readValue("""["a","b","c"]""")

Therefore the Jackson Module for Kotlin will infer the the correct type and you do not need a TypeReference instance.

so your case (slightly renamed and fixed the data class, and JSON):

import com.fasterxml.jackson.module.kotlin.readValue

data class MyData(val a: String, val b: Int)
val JSON = jacksonObjectMapper()  

val jsonStr = """[{"a": "value1", "b": 1}, {"a": "value2", "b": 2}]"""
val myList: List<MyData> = JSON.readValue(jsonStr)

You can also use the form:

val myList = JSON.readValue<List<MyData>>(jsonStr)

Without the import you will have an error because the extension function is not found.

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