Question

首次使用Web集成我在向用户注册系统时遇到问题。我的localhost中有一个api来执行此功能。但问题是输出它的回归。传递数据时参数的类型是否也重要？就好像有一个字段要求用户的号码一样，在HTTP POST中，我是否必须以整数形式发回数据？但我确实注意到BasicNameValuePair不允许这样做。

基本程序如下所示：

public JSONObject registerURL(String fname,String lname,String mail,String password,String number,String sex,int role,String dob,String horoscope,String tob,String pob){
        List<NameValuePair> params = new ArrayList<NameValuePair>();
        params.add(new BasicNameValuePair("first_name", fname));
        params.add(new BasicNameValuePair("last_name", lname));
        params.add(new BasicNameValuePair("email",mail));
        params.add(new BasicNameValuePair("phone",number));
        params.add(new BasicNameValuePair("password",password));
        params.add(new BasicNameValuePair("sex",sex));
        params.add(new BasicNameValuePair("role", String.valueOf(role)));
        params.add(new BasicNameValuePair("date_of_birth",dob));
        params.add(new BasicNameValuePair("horoscope_id",horoscope));
        params.add(new BasicNameValuePair("time_of_birth",tob));
        params.add(new BasicNameValuePair("place_of_birth",pob));
        try {
            json = jsonParser.getJSONFromUrl(registerURL, params, POST);
        } catch (URISyntaxException e) {
            e.printStackTrace();
        }
        return json;
    }

它呼叫的地方

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

        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpResponse httpResponse = null;
        HttpEntity httpentity = null;

        try {
            switch (method) {
            case 1: //PUT
                HttpPut putrequest  = new HttpPut(url);
                putrequest.setEntity(new UrlEncodedFormEntity(params));
                httpResponse  = httpClient.execute(putrequest);
                httpentity = httpResponse.getEntity();
                break;
            case 2: //GET
//              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();          
                break;
            case 3: //POST
                HttpPost postrequest = new HttpPost(url);
                postrequest.setEntity(new UrlEncodedFormEntity(params));
                httpResponse = httpClient.execute(postrequest);
                httpentity = httpResponse.getEntity();

                break;
            }

            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.i("JSON HERE", json);
        } catch (Exception e) {
            Log.e("Buffer Error", "Error converting result " + e.toString());
        }

        // 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 JSON String
        return jObj;

    }

错误如下：

Answer 1

你需要先创建JSONArray而不是JSONObject。因为，你使用了json数组。

JSONArray jsonArray = new JSONArray(jsonString);

JSONObject jsonObject = jsonArray.getJSONObject(index);

在HTTP Post Android上传递参数

1 个答案: