H.N*_*yen 7 serialization kotlin json-deserialization kotlinx.serialization
假设我正在上课:
@Serializable
data class MyClass(
@SerialName("a") val a: String?,
@SerialName("b") val b: String
)
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假设aisnull和b的值为“b 值”,则Json.stringify(MyClass.serializer(), this)产生:
{ "a": null, "b": "b value" }
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基本上如果a是null,我想得到这个:
{ "b": "b value" }
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从一些研究中,我发现这在 Kotlinx Serialization 中是无法开箱即用的,所以我试图构建一个自定义序列化器来显式忽略null值。我从这里开始遵循指南,但无法做出正确的指南。
有人可以照亮我吗?谢谢。
say*_*ame 12
您可以使用explicitNulls = false
例子:
@OptIn(ExperimentalSerializationApi::class)
val format = Json { explicitNulls = false }
@Serializable
data class Project(
val name: String,
val language: String,
val version: String? = "1.3.0",
val website: String?,
)
fun main() {
val data = Project("kotlinx.serialization", "Kotlin", null, null)
val json = format.encodeToString(data)
println(json) // {"name":"kotlinx.serialization","language":"Kotlin"}
}
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https://github.com/Kotlin/kotlinx.serialization/blob/master/docs/json.md#explicit-nulls
试试这个(未经测试,只是基于调整示例):
@Serializable
data class MyClass(val a: String?, val b: String) {
@Serializer(forClass = MyClass::class)
companion object : KSerializer<MyClass> {
override val descriptor: SerialDescriptor = object : SerialClassDescImpl("MyClass") {
init {
addElement("a")
addElement("b")
}
}
override fun serialize(encoder: Encoder, obj: MyClass) {
encoder.beginStructure(descriptor).run {
obj.a?.let { encodeStringElement(descriptor, 0, obj.a) }
encodeStringElement(descriptor, 1, obj.b)
endStructure(descriptor)
}
}
override fun deserialize(decoder: Decoder): MyClass {
var a: String? = null
var b = ""
decoder.beginStructure(descriptor).run {
loop@ while (true) {
when (val i = decodeElementIndex(descriptor)) {
CompositeDecoder.READ_DONE -> break@loop
0 -> a = decodeStringElement(descriptor, i)
1 -> b = decodeStringElement(descriptor, i)
else -> throw SerializationException("Unknown index $i")
}
}
endStructure(descriptor)
}
return MyClass(a, b)
}
}
}
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