Question

我正在开发一个Android应用程序，我遇到了这样一种情况：应用程序将使用API将一些数据发送到php webservice，而webservice将会更新一些json编码的消息，这些消息将被回显。

我的问题是

如何将这个由php echo发送的json消息存储到android app中的变量中？
我如何解析json并使用数据构建一个switch case？

我在某个时候提出了类似的问题并被告知要使用AsyncTask，但我不明白为什么我需要使用它。

将由phpwebservice发送的示例json响应是

{"error":false,"message":"New user created"}

我希望能够获取错误变量并确定是否有任何错误，并在变量中获取消息并将其显示给应用程序中的用户。

我目前有像这样的android signup.java代码

public void post() throws UnsupportedEncodingException
    {
        // Get user defined values
        uname = username.getText().toString();
        email   = mail.getText().toString();
        password   = pass.getText().toString();
        confirmpass   = cpass.getText().toString();
        phone = phn.getText().toString();

        HttpClient httpclient = new DefaultHttpClient();
        HttpResponse httpResponse = null;
        HttpPost httppost = new HttpPost("http://www.rgbpallete.in/led/api/signup");
        if (password.equals(confirmpass)) {
            try {
                List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(4);
                nameValuePairs.add(new BasicNameValuePair("uname", uname));
                nameValuePairs.add(new BasicNameValuePair("pass", password));
                nameValuePairs.add(new BasicNameValuePair("email", email));
                nameValuePairs.add(new BasicNameValuePair("phone", phone));
                httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
                httpResponse = httpclient.execute(httppost);
                //Code to check if user was successfully created
                final int statusCode = httpResponse.getStatusLine().getStatusCode();
                switch (statusCode)
                {
                    case 201:
                        Toast.makeText(getBaseContext(), "Successfully Registered", Toast.LENGTH_SHORT).show();
                        break;
                    case 400:
                        Toast.makeText(getBaseContext(), "Username already taken", Toast.LENGTH_SHORT).show();
                        username.setText("");
                        break;
                    default:
                        Toast.makeText(getBaseContext(), "Unknown error occurred", Toast.LENGTH_SHORT).show();
                }
            } catch (ClientProtocolException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        else
        {
            Toast.makeText(getBaseContext(), "Password mismatch", Toast.LENGTH_SHORT).show();
            //Reset password fields
            pass.setText("");
            cpass.setText("");
        }

    }

虽然这会检查http头代码并且可能有效（我还没有测试过）我想使用jsnon响应并使用它进行处理。

Answer 1

使用java-json：

HttpURLConnection urlConnection = (HttpURLConnection) (uri.toURL().openConnection());
urlConnection.setConnectTimeout(1500);
urlConnection.setRequestMethod("POST");
urlConnection.setDoInput(true);
urlConnection.setDoOutput(true);

List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("uname", uname));
params.add(new BasicNameValuePair("pass", password));
params.add(new BasicNameValuePair("email", email));

OutputStream os = urlConnection.getOutputStream();
BufferedWriter writer = new BufferedWriter(
    new OutputStreamWriter(os, "UTF-8"));
writer.write(getQuery(params));
writer.flush();
writer.close();
os.close();

urlConnection.connect();
if(urlConnection.getResponseCode() == 200){
    InputStream inputStream = new BufferedInputStream(urlConnection.getInputStream());
    BufferedReader streamReader = new BufferedReader(new InputStreamReader(inputStream, "UTF-8"));
    StringBuilder responseStrBuilder = new StringBuilder();

    String inputStr;
    while ((inputStr = streamReader.readLine()) != null)
        responseStrBuilder.append(inputStr);
    JSONObject json = new JSONObject(responseStrBuilder.toString());
    String message = json.getString("message");
}

如何调用php文件并存储json输出

1 个答案: