在 Android 中使用 POST 方法后如何获得响应

Question

网址是：http://localhost 方法类型：post 标头：内容类型：应用程序/json，解码：2 数据：xx

我们如何在 android 中实现这一点？？我如何得到响应？ 我看到 Httpclient 已被弃用

任何帮助将不胜感激

Answer 1

您可以使用 Retrofit 库，因为 Retrofit 比其他方法更快、更容易地从 HTTP 请求加载响应。 访问，https://square.github.io/retrofit/

例如，这里给出的示例 Json 对象，

{
  "title": "foo",
  "body": "bar",
  "userId": 1,
  "id": 101
}  

代码：

public interface APIService {

    @POST("/posts")
    @FormUrlEncoded
    Call<Post> savePost(@Field("title") String title,
                    @Field("body") String body,
                    @Field("userId") long userId);
}

Answer 2

尝试使用 HttpUrlConnection

String Url, query;

    InputStream inputStream;
    HttpURLConnection urlConnection;
    byte[] outputBytes;
    String ResponseData;
    Context context;
try{
            URL url = new URL(Url);
            urlConnection = (HttpURLConnection) url.openConnection();
            outputBytes = query.getBytes("UTF-8");
            urlConnection.setRequestMethod("POST");
            urlConnection.setDoOutput(true);
            urlConnection.setConnectTimeout(15000);
            urlConnection.setRequestProperty("Content-Type", "application/json");
            urlConnection.connect();

            OutputStream os = urlConnection.getOutputStream();
            os.write(outputBytes);
            os.flush();
            os.close();

            inputStream = new BufferedInputStream(urlConnection.getInputStream());
            ResponseData = convertStreamToString(inputStream);

    } catch (MalformedURLException e) {

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

            }


    public String convertStreamToString(InputStream is) {

        BufferedReader reader = new BufferedReader(new InputStreamReader(is));
        StringBuilder sb = new StringBuilder();

        String line = null;
        try {
            while ((line = reader.readLine()) != null) {
                sb.append((line + "\n"));
            }
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try {
                is.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return sb.toString();
    }

Answer 3

 URL url = new URL("http://yoururl.com");
HttpsURLConnection conn = (HttpsURLConnection) url.openConnection();
conn.setReadTimeout(10000);
conn.setConnectTimeout(15000);
conn.setRequestMethod("POST");
conn.setDoInput(true);
conn.setDoOutput(true);
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("firstParam", paramValue1));
params.add(new BasicNameValuePair("secondParam", paramValue2));
params.add(new BasicNameValuePair("thirdParam", paramValue3));
OutputStream os = conn.getOutputStream();
BufferedWriter writer = new BufferedWriter(
        new OutputStreamWriter(os, "UTF-8"));
writer.write(getQuery(params));
writer.flush();
writer.close();
os.close();
conn.connect();

.................................................. ....................

private String getQuery(List<NameValuePair> params) throws UnsupportedEncodingException
{
    StringBuilder result = new StringBuilder();
    boolean first = true;

    for (NameValuePair pair : params)
    {
        if (first)
            first = false;
        else
            result.append("&");

        result.append(URLEncoder.encode(pair.getName(), "UTF-8"));
        result.append("=");
        result.append(URLEncoder.encode(pair.getValue(), "UTF-8"));
    }

    return result.toString();
}

Answer 4

如果您不想使用库，可以使用 HttpURLConnection，否则您也可以使用 Volley 进行网络调用，因为 volley 可以更轻松地管理网络调用！ 谢谢！