JSON解析问题 [英] JSON parsing problem

问题描述

我试图JSON解析在我的Andr​​oid应用程序的链接是<一个href="https://www.buzzador.com/apps/$p$psent_software/webservice/index.php?op=ProductQ&campaign_id=607&userid=10776" rel="nofollow">https://www.buzzador.com/apps/$p$psent_software/webservice/index.php?op=ProductQ&campaign_id=607&userid=10776

当我把它变成JSON对象它给了错误，我 错误是： 08-31 14：40：​​52.281：WARN / System.err的（416）：org.json.JSONException：java.lang.String类型的值不能转换为JSONObject的

 公共静态字符串getmyproductquestiondetails（字符串userid，
        串campaignid）{// https://www.buzzador.com/apps/$p$psent_software/webservice/index.php?op=EducationResult&userid=1&questionid=1,2,3&answergivenbyuser=1,1,0
    字符串数据= NULL;
    尝试 {

        网址URL =新的URL（
                http://dignizant.com/buzz/webservice/index.php?op=getProductQuestion&userid=
                        +的userid +与＆amp; campaign_id =+ campaignid）;
        如果（url.getProtocol（）。与toLowerCase（）。等于（https开头））{
            trustAllHosts（）;
            HttpsURLConnection HTTPS =（HttpsURLConnection）网址
                    .openConnection（）;
            https.setHostnameVerifier（DO_NOT_VERIFY）;
            HTTP = HTTPS;
        } 其他 {
            HTTP =（HttpURLConnection类）url.openConnection（）;
        }
    }赶上（MalformedURLException异常E）{
        // TODO自动生成的catch块
        e.printStackTrace（）;
    }赶上（IOException异常E）{
        // TODO自动生成的catch块
        e.printStackTrace（）;
    }
    utils的utils的=新utils的（）;

    尝试 {
        数据= ut​​ils.convertStreamToString（http.getInputStream（））;
        的System.out.println（getproduct细节响应::+数据）;

    }赶上（IOException异常E）{
        // TODO自动生成的catch块
        e.printStackTrace（）;

        数据= e.toString（）;

    }
    返回的数据;
}







尝试 {


        JSONObject的乔=新的JSONObject（响应）;

    }赶上（例外五）{
        // TODO：处理异常
        e.printStackTrace（）;
    }
 

解决方案

 的char [] UTF8 = NULL;
            StringBuilder的properString =新的StringBuilder（）;

            UTF8 = Response.toCharArray（）;

            的for（int i = 0; I＆LT; utf8.length;我++）{
                如果（（int）的UTF8 [1]  - ; 65000）{
                    properString.append（UTF8 [I]）;
                }
            }
  的System.out.println（登录的响应::
                        + properString.toString（））;
 

i am trying to Json parsing in my android app the link is https://www.buzzador.com/apps/present_software/webservice/index.php?op=ProductQ&campaign_id=607&userid=10776

when i put it into Json object it gives errors to me error is : 08-31 14:40:52.281: WARN/System.err(416): org.json.JSONException: Value of type java.lang.String cannot be converted to JSONObject

public static String getmyproductquestiondetails(String userid,
        String campaignid) {// https://www.buzzador.com/apps/present_software/webservice/index.php?op=EducationResult&userid=1&questionid=1,2,3&answergivenbyuser=1,1,0
    String data = null;
    try {

        URL url = new URL(
                "http://dignizant.com/buzz/webservice/index.php?op=getProductQuestion&userid="
                        + userid + "&campaign_id=" + campaignid);
        if (url.getProtocol().toLowerCase().equals("https")) {
            trustAllHosts();
            HttpsURLConnection https = (HttpsURLConnection) url
                    .openConnection();
            https.setHostnameVerifier(DO_NOT_VERIFY);
            http = https;
        } else {
            http = (HttpURLConnection) url.openConnection();
        }
    } catch (MalformedURLException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    Utils utils = new Utils();

    try {
        data = utils.convertStreamToString(http.getInputStream());
        System.out.println("getproduct details response :: " + data);

    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();

        data = e.toString();

    }
    return data;
}







try {


        JSONObject jo = new JSONObject(response);

    } catch (Exception e) {
        // TODO: handle exception
        e.printStackTrace();
    }

解决方案

  char[] utf8 = null;
            StringBuilder properString = new StringBuilder("");

            utf8 = Response.toCharArray();

            for (int i = 0; i < utf8.length; i++) {
                if ((int) utf8[i] < 65000) {
                    properString.append(utf8[i]);
                }
            }
  System.out.println("Response of Login::"
                        + properString.toString());

这篇关于JSON解析问题的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！