Use Fastjson in Kotlin

Use Fastjson in Kotlin

In Kotlin, we use to make a data class to hold data, and then we can use Fastjson to serialize data-class object to json string or deserialize json string to data-class object now(Since Fastjson 1.2.37).There are three cases：

note: in Kotlin, DataClassName::class, equals DataClassName.class in Java

1.Data Class without Any Annotation

Define DataClassSimple data-class:

data class DataClassSimple(val a : Int, val b : Int)

then, we can use this code below to serialize and deserialize:

    val dts = DataClassSimple(1,2)
    val jsons = JSON.toJSONString(dts)
    println(jsons)
    val clzs = DataClassSimple::class
    println(clzs.javaObjectType)
    val dt2 = JSON.parseObject(jsons,clzs.javaObjectType)
    println(dt2)

the output:

{"a":1,"b":2}
class DataClassSimple
DataClassSimple(a=1, b=2)

2.Data Class with @JSONField Annotation

Define DataClass data-class:

data class DataClass(@JSONField(name="aa")val a : Int, @JSONField(name="bb")val b : Int)

then we use JSON to serialize and deserialize:

    val dt = DataClass(1,2)
    val json = JSON.toJSONString(dt)
    println(json)
    val clz = DataClass::class
    println(clz.javaObjectType)
    val dt1 = JSON.parseObject(json,clz.javaObjectType)
    println(dt1)

the output:

{"aa":1,"bb":2}
class DataClass
DataClass(a=1, b=2)

3.Data Class with @field:JSONField Annotation

Define DataClassField data-class:

data class DataClassField(@field:JSONField(name="aaa")val a : Int, @field:JSONField(name="bbb")val b : Int)

then we use JSON to serialize and deserialize:

    val dtf = DataClassField(1,2)
    val jsonf = JSON.toJSONString(dtf)
    println(jsonf)
    val clzf = DataClassField::class
    println(clzf.javaObjectType)
    val dt3 = JSON.parseObject(jsonf,clzf.javaObjectType)
    println(dt3)

the output:

{"aaa":1,"bbb":2}
class DataClassField
DataClassField(a=1, b=2)