Question

我正在发出一个API请求，该请求返回一些数组值。我需要序列化这些数组值，以便可以将它们分配给它们相应的类属性（即String类型）。

现在，我知道如何使用GSON对列表进行序列化和反序列化，但是使用Retrofit，映射将自动完成。这意味着，如果我的属性的类型为String，则API调用将返回错误“期望字符串但收到数组”。我该如何解决这个问题，以便我可以毫无故障地将它们接收为数组，然后将它们存储为字符串？

我的API响应：

{
"utterances": [{
        "langs": ["eng", "afr", "xho", "zul"],
        "utts": [
            "Have you been here before?",
            "Was u al hier gewees?",
            "Ingaba wakhe weza apha ngaphambili?",
            "Ingabe uke weza lapha ngaphambilini?"
        ],
        "responses": [
            ["Yes", "No"],
            ["Ja", "Nee"],
            ["Ewe", "Hayi"],
            ["Yebo", "Cha"]
        ]
    },
    {
        "langs": ["eng", "afr", "xho", "zul"],
        "utts": [
            "How are you?",
            "Hoe gaan dit met jou?",
            "unjani?",
            "unjani?"
        ],
        "responses": [
            ["Good", "Bad"],
            ["Goed", "sleg"],
            ["ezilungileyo", "ezimbi"],
            ["kuhle", "kubi"]
        ]
    }
]
}

我的UtteranceResponse类：

class UtteranceResponse {

@SerializedName("status")
var status: String? = null

@SerializedName("count")
var count: Int = 0

@SerializedName("utterances")
var utterances: ArrayList<Utterance>? = null
}

我的话语课：

class Utterance: SugarRecord {

@SerializedName ("langs")
var langs: String? = null

@SerializedName ("utts")
var utterances_text: String? = null

var utterances_tts: String? = null

@SerializedName ("responses")
var responses_text: String? = null

constructor(){

}
}

最后是调用函数：

    fun getUtterancesFromWebservice (){
    val apiService = ApiInterface.create()
    val call = apiService.getUtteranceDetails()

    call.enqueue(object: Callback<UtteranceResponse> {
        override fun onResponse(call: Call<UtteranceResponse>, response: retrofit2.Response<UtteranceResponse>?) {
            if (response != null) {
                if (response.body()?.utterances != null){
                    var list: List<Utterance> = response.body()?.utterances!!
                    val utterances: Utterance = list[0]
                    //storeUtterancesFromList(list)
                } else {
                    Log.d ("Response:", response.body().toString())
                }
            }else{
                Log.d ("responseResult", "NULL")
            }

        }

        override fun onFailure(call: Call<UtteranceResponse>, t: Throwable) {
            Log.e("SHIT", t.toString())

        }
    })
}

更新我的API接口也是如此：

@GET("bins/1ahazo")
abstract fun getUtteranceDetails():Call<UtteranceResponse>

    companion object Factory {
        const val BASE_URL = "https://api.myjson.com/"
        fun create(): ApiInterface {
            val gson = GsonBuilder().setPrettyPrinting().create()
            val retrofit = Retrofit.Builder()
                    .baseUrl(BASE_URL)
                    .addConverterFactory(GsonConverterFactory.create())
                    .build()
            return retrofit.create(ApiInterface::class.java)
        }
    }

Answer 1

您将返回单个对象而不是列表。将ApiInterface中的Call<UtteranceResponse>更改为

Call<List<Utterance>>

并将列表转换为字符串list to string and string to list

class Utterance: SugarRecord {

@SerializedName ("langs")
var langs: List<String?>? = null 

@SerializedName ("utts")
var utterances_text: String? = null

var utterances_tts: List<String?>? = null

@SerializedName ("responses")
var responses_tex:List<List<String?>?>? = null;

constructor(){

 }
}

如何使GSON＆Retrofit序列化并将数组从JSON响应映射到字符串？

1 个答案: