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JSON解析问题

我正在解析一个JSON文件(这是有效的).它适用于Android 4.0 - 4.0.4但不适用于较旧的Android版本.这是我的清单的一部分:

<uses-sdk
    android:minSdkVersion="7"
    android:targetSdkVersion="14" />

这是我的解析代码:

public JSONObject getJSONFromUrl(String url) {
    try {
        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost(url);

        HttpResponse httpResponse = httpClient.execute(httpPost);
        HttpEntity httpEntity = httpResponse.getEntity();
        is = httpEntity.getContent();

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    try {
        BufferedReader reader = new BufferedReader(new InputStreamReader(
                is, "UTF-8"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line …

我想在我的Android应用程序中解析Json解析链接是https://www.buzzador.com/apps/present_software/webservice/index.php?op=ProductQ&campaign_id=607&userid=10776

当我把它放入Json对象时它会给我带来错误错误是:08-31 14:40:52.281:WARN/System.err(416):org.json.JSONException:java.lang.String类型的值无法转换到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 …

我正在尝试使用从Web服务接收的以下字符串初始化JSONObject:

"{
    "campaignid": "8",
    "campaignname": "Pilotarienak 2011",
    "campaignlink": "http:\\/\\/www.xxx.com\\/fr\\/cote-basque\\/agenda\\/2011-05-20\\/FMAAQU064FS016DV-pilotarienak-d-anglet?fromapp",
    "splash": "http:\\/\\/www.xxx.com\\/ads\\/customers\\/pilotarienak\\/320x480.jpg",
    "banner": "http:\\/\\/www.xxx.com\\/ads\\/customers\\/pilotarienak\\/320x160.jpg"
}"

它似乎是有效的json(它在jsonlint.com中验证),但是在初始化JSONObject时,我得到:

org.json.JSONException: Value  of type java.lang.String cannot be converted to JSONObject

有人可以帮忙吗？

谢谢