Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
914 views
in Technique[技术] by (71.8m points)

json - 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?
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...