Question

我需要从php server获取listview json。我的api存在问题，这是我的api链接

http://staging.homeneedsonline.com/ws/ws_exe_newenq.php?exeId=2

在我的浏览器中，我得到了适当的回复。在android asyn任务中这个相同的url我得到错误响应。

{"success":"0","message":"New enquiry not found ."}

可能这将是无逻辑的问题，但我找不到错误。这个Api问题或应用程序问题？我做错了什么？

这是我的代码

 protected JSONObject doInBackground(String... args) {
              // Building Parameters
              List<NameValuePair> params = new ArrayList<NameValuePair>();

              UserFunctions userFunctions = new UserFunctions();
              HashMap<String, String> user = userFunctions.getExinfo(EnquiryActivity.this);

             String userid1 = user.get("exid");
               System.out.println(userid1);


              final String URL_LIST1 = "http://staging.homeneedsonline.com/ws/ws_exe_newenq.php?exeId=2";


                 System.out.println(URL_LIST1);  
              JSONParser jsonParser = new JSONParser();    

              JSONObject json = jsonParser.getJSONFromUrl(URL_LIST1, get, params);
              System.out.println(json);

getJSONFROmUrl方法

public JSONObject getJSONFromUrl(String url,String method, List<NameValuePair> params) {

        // Making HTTP request
        try {
            if (method == "POST") {
                // request method is POST
                // defaultHttpClient
                DefaultHttpClient httpClient = new DefaultHttpClient();
                HttpPost httpPost = new HttpPost(url);
                httpPost.setEntity(new UrlEncodedFormEntity(params));

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

            } else if (method == "GET") {
                // request method is GET
                DefaultHttpClient httpClient = new DefaultHttpClient();
                String paramString = URLEncodedUtils.format(params, "utf-8");
                url += "?" + paramString;
                HttpGet httpGet = new HttpGet(url);
                httpGet.addHeader("Cache-Control", "no-cache");
                HttpResponse httpResponse = httpClient.execute(httpGet);
                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, "iso-8859-1"), 8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) {
                sb.append(line + "\n");
            }
            is.close();
            json = sb.toString();
            Log.e("JSON", json);
        } catch (Exception e) {
            Log.e("Buffer Error", "Error converting result " + e.toString());
            return null;
        }

        // try parse the string to a JSON object
        try {
            jObj = new JSONObject(json);           
        } catch (JSONException e) {
            Log.e("JSON Parser", "Error parsing data " + e.toString());
            return null;

        }

        // return JSON String
        return jObj;

    }

Answer 1

你传递了一个params地图，你的getJSONfromUrl附加了一个“？”到您的URL后跟这些参数。但是，在您的特定情况下，参数是空的，并且您实际上已经在url字符串中提供了查询参数。这意味着您实际获得的URL是 http://staging.homeneedsonline.com/ws/ws_exe_newenq.php?exeId=2？

请注意尾随问号。如果您在浏览器中打开此URL，则会收到相同的错误响应。

解决方案是

从URL中删除exeId参数，然后将其作为参数映射的一部分传递给getJSONfromUrl追加
完全停止传递参数并假设网址已经完成（在您的情况下）
在向网址添加任何查询之前，添加一项检查以查看param地图是否为空（这应该在任何情况下完成）

if (params.size() > 0) { String paramString = URLEncodedUtils.format(params, "utf-8"); url += "?" + paramString; }

另请注意，您应使用String.equals()而不是==来比较您的“GET”，“POST”等字符串。

Answer 2

像这样更改您的JSONParser班级GET。

           else if (method == "GET") {
                // request method is GET
                DefaultHttpClient httpClient = new DefaultHttpClient();
                String paramString = URLEncodedUtils.format(params, "utf-8");
                // url += "?" + paramString;
                HttpGet httpGet = new HttpGet(url);
                httpGet.addHeader("Cache-Control", "no-cache");
                HttpResponse httpResponse = httpClient.execute(httpGet);
                HttpEntity httpEntity = httpResponse.getEntity();
                is = httpEntity.getContent();
            }

原因是不想要的空白＆＃34; params＆＃34;通过你的URL传递，没有必要   传递＆＃34; params＆＃34;如果没有必要。

如果＆＃34; params＆＃34;为null，它会像下面一样更改您的网址。

http://staging.homeneedsonline.com/ws/ws_exe_newenq.php?exeId=2？

此网址向您提供了错误回复。

因此，请避免追加＆＃34; params＆＃34;在网址中。

Android中的JSON结果错误

2 个答案: