Question

我知道这是旧的或重复的问题。但我没有弄到错的地方。我从这个JSON得到了结果null。

{
"error": {
    "dateTimeUtc": "2017-12-26T05:46:05.1126801+00:00",
    "errorReference": "sample string 2",
    "errorType": "Error",
    "title": "sample string 3",
    "code": "sample string 4",
    "messages": [
      "sample string 1",
      "sample string 2"
    ]
  },
  "result": {
    "message": "sample string 1"
  }
}

我有从Retrofit2读取的这种类型的JSON。我创建了一个名为 RetrofitResponce

的POJO类

private Error error;
private Result result;
public Result getResult ()
{
    return result;
}

public void setResult (Result result)
{
    this.result = result;
}

public Error getError ()
{
    return error;
}

public void setError (Error error)
{
    this.error = error;
}

@Override
public String toString()
{
    return "ClassPojo [result = "+result+", error = "+error+"]";
}`

现在在call.enque（）方法中处理响应，比如

call.enqueue(new Callback<RegisterUser>() {
        @Override
        public void onResponse(Call<RetrofitResponce> call, Response<RetrofitResponce> response) {

            try {
                RetrofitResponce retrofitResponce= response.body();

                Error error = retrofitResponce.getError();
                Result result=retrofitResponce.getResult();


                Log.e("Eroor", "rr  " + error.getTitle().toString());
            } catch (Exception e) {

            }
        }

        @Override
        public void onFailure(Call<RetrofitResponce> call, Throwable t) {
            Log.e("Failure ", "fail " + t.toString());
        }
    });

但是在这里我得到了NullPointerException，即。我在RestrofitResponce中没有得到任何回复。错误和结果都是空的。

java.lang.NullPointerException：尝试调用虚方法＆net;来自net.xyc.abc.Model.Model.RegisterUser.getError（）＆＃39;在空对象引用上

请帮帮我。试着解决我的问题。提前谢谢。

PostMan回复，

{
"error": {
    "dateTimeUtc": "2017-12-26T07:05:51.1427712Z",
    "errorReference": "00000000-0000-0000-0000-000000000000",
    "errorType": "Error",
    "title": "The request is invalid.",
    "code": null,
    "messages": [
        "Passwords must have at least one non letter or digit character. Passwords must have at least one uppercase ('A'-'Z')."
    ]
},
"result": null

}

Answer 1

您必须在Response类中为每个与Web服务名称相同的字段添加@SerializedName。

e.g

@SerializedName("error")
private Error error;
@SerializedName("result")
private Result result;

请使用此网站从JSON Response创建您的课程。 http://www.jsonschema2pojo.org/

之后你必须检查是否成功回应。

call.enqueue(new Callback<RegisterUser>() {
        @Override
        public void onResponse(Call<RetrofitResponce> call, Response<RetrofitResponce> response) {

            try 
            {
                if(response.isSuccessful() && response.body() != null)
                {
                    RetrofitResponce retrofitResponce= response.body();
                    Error error = retrofitResponce.getError();
                    Result result=retrofitResponce.getResult();
                 }

            } catch (Exception e) {

            }
        }

        @Override
        public void onFailure(Call<RetrofitResponce> call, Throwable t) {
            Log.e("Failure ", "fail " + t.toString());
        }
    });

Answer 2

改造回调需要响应模型，只需更改：

str1 <- "23-11-17 12:53:41 PM"

快乐的编码!!

Answer 3

为RetrofitResponce

尝试此Pojo

public class RetrofitResponce {

@SerializedName("error")
@Expose
private Error error;
@SerializedName("result")
@Expose
private Result result;

public Error getError() {
return error;
}

public void setError(Error error) {
this.error = error;
}

public Result getResult() {
return result;
}

public void setResult(Result result) {
this.result = result;
}

}


public class Result {

@SerializedName("message")
@Expose
private String message;

public String getMessage() {
return message;
}

public void setMessage(String message) {
this.message = message;
}

}


public class Error {

@SerializedName("dateTimeUtc")
@Expose
private String dateTimeUtc;
@SerializedName("errorReference")
@Expose
private String errorReference;
@SerializedName("errorType")
@Expose
private String errorType;
@SerializedName("title")
@Expose
private String title;
@SerializedName("code")
@Expose
private String code;
@SerializedName("messages")
@Expose
private List<String> messages = null;

public String getDateTimeUtc() {
return dateTimeUtc;
}

public void setDateTimeUtc(String dateTimeUtc) {
this.dateTimeUtc = dateTimeUtc;
}

public String getErrorReference() {
return errorReference;
}

public void setErrorReference(String errorReference) {
this.errorReference = errorReference;
}

public String getErrorType() {
return errorType;
}

public void setErrorType(String errorType) {
this.errorType = errorType;
}

public String getTitle() {
return title;
}

public void setTitle(String title) {
this.title = title;
}

public String getCode() {
return code;
}

public void setCode(String code) {
this.code = code;
}

public List<String> getMessages() {
return messages;
}

public void setMessages(List<String> messages) {
this.messages = messages;
}

}

Answer 4

我相信这会对您有所帮助。

您无法通过Single模型处理Success and Faille。要获取错误正文，您需要调用response.errorBody（） .string （）

尝试以下代码：

call.enqueue(new Callback<RegisterUser>() {
    @Override
    public void onResponse(Call<RetrofitResponce> call, Response<RetrofitResponce> response) {

       try {
                if (response.isSuccessful()) {
                Log.d("Successful",response.body();); 
                } else {
                    Log.d("errorBody ", response.errorBody().string());
                }
            } catch (Exception e) {
                Log.d("Exception ",e.toString());
            }

    }

    @Override
    public void onFailure(Call<RetrofitResponce> call, Throwable t) {
        Log.e("Failure ", "fail " + t.toString());
    }
});

Answer 5

试试这个。

ffill

在改造2.0中处理JSON响应

5 个答案: