如何在Android中获得JSON响应？

我想为我的应用程序获取json数据

下面是获得响应的

CallAPI.java

类

public class CallAPI extends AsyncTask<String,String,String> {

    String jsonStr="";
    private static final String url="https://api.litzscore.com/rest/v2/recent_matches/";

    @Override
    protected String doInBackground(String... params) {
        List<NameValuePair> param=new ArrayList<>();
        param.add(new BasicNameValuePair("access_token",""));//here i am added my access token
        ServiceHandler sh=new ServiceHandler();
        jsonStr=sh.makeServiceCall(url,ServiceHandler.GET,param);

        try{
            JSONObject obj=new JSONObject(jsonStr);//I got error here

            Log.e("response",""+obj.toString());
        }
        catch (JSONException e){
            e.printStackTrace();
        }
        return null;
    }
}

这是我的错误

org.json.JSONException: Value ���nҩV�͗Qo�0ǿ��ץ`�o���j+� of type java.lang.String cannot be converted to JSONObject

---

如果我从本地服务器（如wammp、xampp）调用该文件，那么它工作正常。

尝试像这样转换json响应

public class ServiceHandler {
static String response=null;
InputStream is = null;
public final static int GET = 1;
public final static int POST = 2;

public ServiceHandler() {

}

public String makeServiceCall(String url, int method) {
    return this.makeServiceCall(url, method, null);
}
public String makeServiceCall(String url, int method,
                              List<NameValuePair> params) {
    try {
        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpEntity httpEntity = null;
        HttpResponse httpResponse = null;

        if (method == POST) {
            HttpPost httpPost = new HttpPost(url);
            if (params != null) {
                httpPost.setEntity(new UrlEncodedFormEntity(params));
            }

            httpResponse = httpClient.execute(httpPost);

        } else if (method == GET) {
            if (params != null) {
                String paramString = URLEncodedUtils
                        .format(params, "utf-8");
                url += "?" + paramString;
            }
            Log.e("URL",""+url);
            HttpGet httpGet = new HttpGet(url);

            httpResponse = httpClient.execute(httpGet);
        }
        httpEntity = httpResponse.getEntity();
        is = httpEntity.getContent();

        //convert response to string
        BufferedReader reader = new BufferedReader(new InputStreamReader(is, "iso-8859-1"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");

        }
        is.close();
        response = sb.toString();

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }
    Log.e("RESPONSE",response);
    return response;

}

}

公共类ServiceHandler{
静态字符串响应=null；
InputStream=null；
公共最终静态int GET=1；
公共最终静态int POST=2；
公共服务处理程序（）{
}
公共字符串makeServiceCall（字符串url，int方法）{
返回此.makeServiceCall（url，方法，null）；
}
公共字符串makeServiceCall（字符串url，int方法，
列表参数）{
试一试{
DefaultHttpClient httpClient=新的DefaultHttpClient（）；
HttpEntity HttpEntity=null；
HttpResponse HttpResponse=null；
if（方法==POST）{
HttpPost HttpPost=新的HttpPost（url）；
如果（参数！=null）{
setEntity（新的UrlEncodedFormEntity（参数））；
}
httpResponse=httpClient.execute（httpPost）；
}else if（方法==GET）{
如果（参数！=null）{
String paramString=URLEncodedUtils
.格式（参数“utf-8”）；
url+=“？”+参数字符串；
}
Log.e（“URL”，“URL+URL”）；
HttpGet HttpGet=新的HttpGet（url）；
httpResponse=httpClient.execute（httpGet）；
}
httpEntity=httpResponse.getEntity（）；
is=httpEntity.getContent（）；
//将响应转换为字符串
BufferedReader reader=新的BufferedReader（新的InputStreamReader（is，“iso-8859-1”），8；
StringBuilder sb=新的StringBuilder（）；
字符串行=null；
而（（line=reader.readLine（））！=null）{
sb.追加（第+行“\n”）；
}
is.close（）；
response=sb.toString（）；
}捕获（不支持的编码异常e）{
e、 printStackTrace（）；
}捕获（客户端协议例外e）{
e、 printStackTrace（）；
}捕获（IOE异常）{
e、 printStackTrace（）；
}
Log.e（“响应”，响应）；
返回响应；
}
}

���nҩV�͗Qo�0ǿ��ץ

�o���j+�` 无效的JSON字符串这就是问题所在，我该怎么办？在

Log.e（“RESPONSE”，RESPONSE）中从服务器获取什么信息行？在响应日志中记录���nҩV�͗Qo�0ǿ��ץ`�o��......... 就像你们在浏览器上检查api的响应或者其他什么，它会重新返回相同的
���nҩV�͗Qo�0ǿ��ץ`�o���j+�？不，它不起作用，因为我的服务提供商提供了gzip格式的数据，所以我需要解压缩这些数据。这不是你的错。@MilanGajera我认为你应该为你的课程添加更多细节。例如：关于服务器的响应格式。这很容易，因为每个人都能帮助你。我得到了这样的回应���nҩV�͗Qo�0ǿ��ץ`�o���j+�我的意思是，编辑您的问题，并添加关于为什么您的服务器返回gzip格式而不是json格式的信息。显然，找到解决这个问题的办法是必要的。
public class ServiceHandler {
static String response=null;
InputStream is = null;
public final static int GET = 1;
public final static int POST = 2;

public ServiceHandler() {

}

public String makeServiceCall(String url, int method) {
    return this.makeServiceCall(url, method, null);
}
public String makeServiceCall(String url, int method,
                              List<NameValuePair> params) {
    try {
        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpEntity httpEntity = null;
        HttpResponse httpResponse = null;

        if (method == POST) {
            HttpPost httpPost = new HttpPost(url);
            if (params != null) {
                httpPost.setEntity(new UrlEncodedFormEntity(params));
            }

            httpResponse = httpClient.execute(httpPost);

        } else if (method == GET) {
            if (params != null) {
                String paramString = URLEncodedUtils
                        .format(params, "utf-8");
                url += "?" + paramString;
            }
            Log.e("URL",""+url);
            HttpGet httpGet = new HttpGet(url);

            httpResponse = httpClient.execute(httpGet);
        }
        httpEntity = httpResponse.getEntity();
        is = httpEntity.getContent();

        //convert response to string
        BufferedReader reader = new BufferedReader(new InputStreamReader(is, "iso-8859-1"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");

        }
        is.close();
        response = sb.toString();

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }
    Log.e("RESPONSE",response);
    return response;

}

}