Gson - 如何在 Kotlin 中解析部分 json

Question

我有一个data class，叫做Model，

data class Model(var name: String? = null, var address: String? = null) {

    override fun toString(): String {
        return "name: $name address: $address"
    }
}

我有两个 json 字符串，

val json1 = "{ \"name\": \"Alex\", \"address\": \"rome, 1000\" }"
val json2 = "{ \"name\": \"Alex\", \"address\": {\"city\": \"rome\", \"post\": \"1000\" } }"

下面的转换工作正常，

val model1 = Gson().fromJson<Model>(json1, Model::class.java)

但是，这个版本不工作。给出一个例外。

val model2 = Gson().fromJson<Model>(json2, Model::class.java)

例外：

E/AndroidRuntime: FATAL EXCEPTION: main
Process: com.achellies.kotlin, PID: 11211
java.lang.RuntimeException: Unable to start activity ComponentInfo{com.achellies.kotlin/com.achellies.kotlin.MainActivity}: com.google.gson.JsonSyntaxException: com.google.gson.stream.MalformedJsonException: Unterminated object at line 1 column 40 path $.address
    at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:3086)
    at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:3229)
    at android.app.servertransaction.LaunchActivityItem.execute(LaunchActivityItem.java:78)
    at android.app.servertransaction.TransactionExecutor.executeCallbacks(TransactionExecutor.java:108)
    at android.app.servertransaction.TransactionExecutor.execute(TransactionExecutor.java:68)
    at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1926)
    at android.os.Handler.dispatchMessage(Handler.java:106)
    at android.os.Looper.loop(Looper.java:214)
    at

如何解析json2 以便address 字段将保存内部地址json 数据的字符串值？例如解析后address应该等于

{"city": "rome", "post": "1000"}

Answer 1

你试过Any吗？

data class Model(var name: String? = null, var address: Any? = null) {

    override fun toString(): String {
        return "name: $name address: $address"
    }
}

试试这个，它会工作的。

祝你有美好的一天。 :)

Answer 2

您可以使用专门用于您的 Model 类的 JsonDeserializer，如下所示：

fun main() {
    val json1 = "{ \"name\": \"Alex\", \"address\": \"rome, 1000\" }"
    val json2 = "{ \"name\": \"Alex\", \"address\": {\"city\": \"rome\", \"post\": \"1000\" } }"

    val gson = GsonBuilder().registerTypeAdapter(Model::class.java, ModelDeserializer()).create()

    println(gson.fromJson(json1, Model::class.java))
    println(gson.fromJson(json2, Model::class.java))
}

data class Model(val name: String, val address: String)

class ModelDeserializer : JsonDeserializer<Model> {

    override fun deserialize(json: JsonElement, typeOfT: Type, context:     JsonDeserializationContext): Model {
        json as JsonObject

        val name = json.get("name").asString

        val addressJson = json.get("address")
        val address = if (addressJson.isJsonObject) addressJson.asJsonObject.toString() else addressJson.asString

        return Model(name, address)
    }
}

哪些输出：

Model(name=Alex, address=rome, 1000)
Model(name=Alex, address={"city":"rome","post":"1000"})

如果你使用 Kotson 库，看起来你就是这样，事情会变得更好：

val gson = GsonBuilder()
        .registerTypeAdapter<Model> {
            deserialize {
                val json = it.json.asJsonObject

                val name = json["name"].asString

                val addressJson = json["address"]
                val address = if (addressJson.isJsonObject) addressJson.asJsonObject.toString() else addressJson.asString

                Model(name, address)
            }
        }
        .create()

println(gson.fromJson<Model>(json1))
println(gson.fromJson<Model>(json2))

Answer 3

请尝试Any，它应该可以工作。

var address: String? = null

Answer 4

如果地址本身是一个有效的 json，那么你可以这样做：

val jsonObject = JSONObject(jsonString)
val address = jsonObject.optJSONObject("address").toString()

如果是字符串则

val address = jsonObject.optString("address").toString()