android - 通过 GET 方法发送值

Question

我一直在尝试通过 GET 方法将数据发送到服务器，但我无法找到实现它的方法。我在异步任务中尝试了一些代码，但没有。 Web 服务是用 cakePhp 制作的，格式如下:

Base_URI/users/add.json?json={“email”: xxx@x.com, “password”: “xxxxxxxxx”, “first_name”: “Xyz”, “last_name”: “Xyz”}

请 Android 专家找出解决此问题的方法。谢谢

代码如下:

List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
            nameValuePairs.add(new BasicNameValuePair("email", UserDummy.email));
            nameValuePairs.add(new BasicNameValuePair("password", UserDummy.password));
            nameValuePairs.add(new BasicNameValuePair("first_name", UserDummy.fname));
            nameValuePairs.add(new BasicNameValuePair("last_name", UserDummy.lname));
            // Making HTTP request
    try {

        // check for request method
        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, HTTP.UTF_8);
            url += "?json={" + paramString+"}";                                                                                                                                                     ;
            HttpGet httpGet = new HttpGet(url);

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

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
        Log.v("XXX", e.getMessage());
    } catch (ClientProtocolException e) {
        e.printStackTrace();
        Log.v("XXX", e.getMessage());
    } catch (IOException e) {
        e.printStackTrace();
        Log.v("XXX", e.getMessage());
    }

    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();
    } 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;

HttpGet 不接受这种格式的 url 并给出了错误，但是当我在浏览器上尝试它时它工作正常。错误如下:

Illegal character in query at index 56

Answer 1

终于在一整天的调试和尝试不同的解决方案之后，我解决了这个问题:)

我需要像这样对参数部分而不是整个 URL 进行编码:

String url = "Base_URI/users/add.json?json=";
    url =url +  URLEncoder.encode("{\"email\":\""+email+"\",\"password\":\""+password+"\"}", "UTF-8");

感谢大家的支持!