Question

我想获取应用程序的json数据。

以下是获得回复的CallAPI.java课程

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json ServiceHandler.java

public class ServiceHandler {
    static String response=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();
            response = EntityUtils.toString(httpEntity);

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

    }

}

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调用文件，如wammp，xampp，那么它正常工作。请帮我解决这个问题。

Answer 1

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;

}

}

我怎样才能在android中获得JSON响应？

1 个答案: