Android 如何从一个JSON响应中分割两个对象的数据？

我从服务器得到json响应…但它包含两个对象的数据…一个是ArrayList类型，第二个是POJO（HomeVO）类。我想将数据拆分并存储到不同的对象中。我是usnig GSON api

Servlet:

        response.setContentType("application/json");
        response.setCharacterEncoding("UTF-8");
        response.getWriter().write(new Gson().toJson(questions));
        response.getWriter().write(new Gson().toJson(homeVo));


 Json Response:

[{"questionId":2,"question":"Quality","typeOfQuestion":2},    {"questionId":3,"question":"Satisfaction","typeOfQuestion":1},{"questionId":4,"question":"overall","typeOfQuestion":2}]{"feedbackName":"IMS","expiryDate":"2014-12-12","createdDate":"2014-10-24","feedbackId":2}



Android Parsing:



HttpClient httpClient = WebServiceUtils.getHttpClient();
        try {
            HttpResponse response = httpClient.execute(new HttpGet(url));
            HttpEntity entity = response.getEntity();
            Reader reader = new InputStreamReader(entity.getContent());

            data = gson.fromJson(reader, arrayListType);
        } catch (Exception e) {
            e.printStackTrace();
            Log.i("json array",
                    "While getting server response server generate error. ");
        }

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我相信你们有两个POJO的问题和HomeVO课程。然后按照以下步骤操作：

您可以使用两个列表（问题和homeVo）创建另一个DTO

公共类ResultDTO{
私人列表homeVoList；
私人列表<问题>问题列表；
//在此处使用您的getter和setter：
}

您有两种选择： 1.手动解析字符串（不推荐使用） 2.使用Gson将JSon对象转换为对象，然后使用Gson将其转换回一个JSon对象

如果你需要更详细的信息，请告诉我

更多说明：

假设您有两个不同的JSon字符串，称为JsonA和JSonB。 为了加入他们，你必须下载Gson库

class AClass{
int idA;
String nameA;
} // Note that the variable's names must be the same as the identifiers in JSON

class BClass{
int idB;
String nameB;
}
class JoinedClass{
BClass bClass;
AClass aClass; //don't forget getters and setters
}
public String joinJson(String JsonA , String JsonB){
Gson gson = new Gson();
AClass aClass = new AClass();
BClass bClass = new BClass();

aClass = gson.fromJson(JsonA, AClass.class);
bClass = gson.fromJson(JsonB, BClass.class);
JoinedClass joinedClass = new JoinedClass();
joinedClass.setAClass(aClass );
joinedClass.setBClass(bClass);
return gson.toJson(joinedClass);
}
// but you know, just after writing this code, i found that there might be an easier way to do this. 
// Thanks for attention!

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您的json无效。您最好先更改servlet端。您没有返回有效的json。您将两个JSON并排连接。是的，请详细说明您的答案…thnx