Android - POST到RESTful Web服务

时间：2015-11-16 23:40:53

标签： android rest post

我正在寻找有关如何在我的Android应用程序中将数据发布到Web服务的一些指导。不幸的是，这是一个学校项目，所以我无法使用外部库。

Web服务有一个基本URL，例如：

http://example.com/service/create

并采用以下格式的两个变量：

username = "user1"
locationname = "location1"

Web服务是RESTful并使用XML结构，如果这有所不同。从我的研究中我了解到我应该使用URLconnection而不是弃用的HTTPconnection，但我找不到我要找的例子。

这是我的尝试，目前无效：

import android.os.AsyncTask;
import android.os.Bundle;
import android.support.v7.app.AppCompatActivity;

import java.io.BufferedOutputStream;
import java.io.IOException;
import java.io.OutputStream;
import java.net.HttpURLConnection;
import java.net.ProtocolException;
import java.net.URL;

public class MainActivity extends AppCompatActivity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        toPost test = new toPost();
        text.execute();
    }

    private class toPost extends AsyncTask<URL, Void, String> {
        @Override
        protected String doInBackground(URL... params) {
            HttpURLConnection conn = null;
            try {
                URL url = new URL("http://example.com/service");
                conn = (HttpURLConnection) url.openConnection();
                conn.setReadTimeout(10000);
                conn.setConnectTimeout(15000);
                conn.setRequestMethod("POST");
                conn.setDoInput(true);
                conn.setDoOutput(true);
                String body = "username=user1&locationname=location1";
                OutputStream output = new BufferedOutputStream(conn.getOutputStream());
                output.write(body.getBytes());
                output.flush();
            } catch (ProtocolException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            } finally {
                conn.disconnect();
            }
            return null;
        }
    }

}

7 个答案:

答案 0 :(得分：2)

我会使用排球库，as suggested by google

在该页面上解释了发出请求，但其简单如下：

final TextView mTextView = (TextView) findViewById(R.id.text);
...

// Instantiate the RequestQueue.
RequestQueue queue = Volley.newRequestQueue(this);
String url ="http://www.google.com";

// Request a string response from the provided URL.
StringRequest stringRequest = new StringRequest(Request.Method.GET, url,
            new Response.Listener<String>() {
    @Override
    public void onResponse(String response) {
        // Display the first 500 characters of the response string.
        mTextView.setText("Response is: "+ response.substring(0,500));
    }
}, new Response.ErrorListener() {
    @Override
    public void onErrorResponse(VolleyError error) {
        mTextView.setText("That didn't work!");
    }
});
// Add the request to the RequestQueue.
queue.add(stringRequest);

答案 1 :(得分：2)

如果要使用HttpUrlConnection，可以参考以下两个示例。希望这有帮助！

private class LoginRequest extends AsyncTask<Void, Void, String> {

    @Override
    protected String doInBackground(Void... voids) {
        String address = "http://server/login";
        HttpURLConnection urlConnection;
        String requestBody;
        Uri.Builder builder = new Uri.Builder();
        Map<String, String> params = new HashMap<>();            
        params.put("username", "bnk");
        params.put("password", "bnk123");

        // encode parameters
        Iterator entries = params.entrySet().iterator();
        while (entries.hasNext()) {
            Map.Entry entry = (Map.Entry) entries.next();
            builder.appendQueryParameter(entry.getKey().toString(), entry.getValue().toString());
            entries.remove();
        }
        requestBody = builder.build().getEncodedQuery();

        try {
            URL url = new URL(address);
            urlConnection = (HttpURLConnection) url.openConnection();
            urlConnection.setDoOutput(true);
            urlConnection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
            OutputStream outputStream = new BufferedOutputStream(urlConnection.getOutputStream());
            BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(outputStream, "utf-8"));
            writer.write(requestBody);
            writer.flush();
            writer.close();
            outputStream.close();

            JSONObject jsonObject = new JSONObject();
            InputStream inputStream;
            // get stream
            if (urlConnection.getResponseCode() < HttpURLConnection.HTTP_BAD_REQUEST) {
                inputStream = urlConnection.getInputStream();
            } else {
                inputStream = urlConnection.getErrorStream();
            }
            // parse stream
            BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
            String temp, response = "";
            while ((temp = bufferedReader.readLine()) != null) {
                response += temp;
            }
            // put into JSONObject
            jsonObject.put("Content", response);
            jsonObject.put("Message", urlConnection.getResponseMessage());
            jsonObject.put("Length", urlConnection.getContentLength());
            jsonObject.put("Type", urlConnection.getContentType());

            return jsonObject.toString();
        } catch (IOException | JSONException e) {
            return e.toString();
        }
    }

    @Override
    protected void onPostExecute(String result) {
        super.onPostExecute(result);
        Log.i(LOG_TAG, "POST\n" + result);
    }
}


private class JsonPostRequest extends AsyncTask<Void, Void, String> {

    @Override
    protected String doInBackground(Void... voids) {
        try {
            String address = "http://server/postvalue";
            JSONObject json = new JSONObject();                
            json.put("Title", "Dummy Title");
            json.put("Author", "Dummy Author");
            String requestBody = json.toString();
            URL url = new URL(address);
            HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
            urlConnection.setDoOutput(true);
            urlConnection.setRequestProperty("Content-Type", "application/json");
            OutputStream outputStream = new BufferedOutputStream(urlConnection.getOutputStream());
            BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(outputStream, "utf-8"));
            writer.write(requestBody);
            writer.flush();
            writer.close();
            outputStream.close();

            InputStream inputStream;
            // get stream
            if (urlConnection.getResponseCode() < HttpURLConnection.HTTP_BAD_REQUEST) {
                inputStream = urlConnection.getInputStream();
            } else {
                inputStream = urlConnection.getErrorStream();
            }
            // parse stream
            BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
            String temp, response = "";
            while ((temp = bufferedReader.readLine()) != null) {
                response += temp;
            }
            // put into JSONObject
            JSONObject jsonObject = new JSONObject();
            jsonObject.put("Content", response);
            jsonObject.put("Message", urlConnection.getResponseMessage());
            jsonObject.put("Length", urlConnection.getContentLength());
            jsonObject.put("Type", urlConnection.getContentType());
            return jsonObject.toString();
        } catch (IOException | JSONException e) {
            return e.toString();
        }
    }

    @Override
    protected void onPostExecute(String result) {
        super.onPostExecute(result);
        Log.i(LOG_TAG, "POST RESPONSE: " + result);
        mTextView.setText(result);
    }
}

答案 2 :(得分：1)

是的，你应该使用URLConnection来发出请求。

您可以将xml数据作为有效负载发送。

请参阅Android - Using HttpURLConnection to POST XML data 

    URL url = new URL(url);
    HttpURLConnection conn = (HttpURLConnection) url.openConnection();
    try {
        conn.setReadTimeout(10000);
        conn.setConnectTimeout(15000);
        conn.setRequestMethod("POST");
        conn.setDoInput(true);
        conn.setDoOutput(true);
        String body = "<xml...</xml>";
        OutputStream output = new BufferedOutputStream(conn.getOutputStream());
        output.write(body.getBytes());
        output.flush();
    } finally {
        conn.disconnect();
    }

答案 3 :(得分：1)

我的建议是在Retrofit之间使用Jackson converter。

Retrofit支持异步和同步请求。它支持GET，POST，PUT，DELETE和HEAD方法。

Jackson将帮助您解析XML到JSON对象。

这两个都非常易于使用并且具有良好的文档。

Here你可以找到使用Retrofit的简单教程。

答案 4 :(得分：1)

使用以下代码从您的Android应用程序调用休息Web服务。这是经过全面测试的代码。

public class RestClient {

    static Context context;
    private static int responseCode;
    private static String response;

    public static String getResponse() {
        return response;
    }

    public static void setResponse(String response) {
        RestClient.response = response;
    }

    public static int getResponseCode() {
        return responseCode;
    }

    public static void setResponseCode(int responseCode) {
        RestClient.responseCode = responseCode;
    }

    public static void Execute(String requestMethod, String jsonData, String urlMethod, Context contextTemp, HashMap<String, Object> params) {
        try {
            context = contextTemp;
            String ip = context.getResources().getString(R.string.ip);
            StringBuilder urlString = new StringBuilder(ip + urlMethod);
            if (params != null) {
                for (Map.Entry<String, Object> para : params.entrySet()) {
                    if (para.getValue() instanceof Long) {
                        urlString.append("?" + para.getKey() + "=" +(Long)para.getValue());
                    }
                    if (para.getValue() instanceof String) {
                        urlString.append("?" + para.getKey() + "=" +String.valueOf(para.getValue()));
                    }
                }
            }

            URL url = new URL(urlString.toString());

            HttpURLConnection conn = (HttpURLConnection) url.openConnection();
            conn.setRequestMethod(requestMethod);
            conn.setReadTimeout(10000 /*milliseconds*/);
            conn.setConnectTimeout(15000 /* milliseconds */);


            switch (requestMethod) {
                case "POST" : case "PUT":
                    conn.setDoInput(true);
                    conn.setDoOutput(true);
                    conn.setRequestProperty("Content-Type", "application/json;charset=utf-8");
                    conn.setRequestProperty("X-Requested-With", "XMLHttpRequest");
                    conn.connect();
                    OutputStream os = new BufferedOutputStream(conn.getOutputStream());
                    os.write(jsonData.getBytes());
                    os.flush();
                    responseCode = conn.getResponseCode();
                    break;
                case "GET":
                    responseCode = conn.getResponseCode();
                    System.out.println("GET Response Code :: " + responseCode);
                    break;
                 break;

            }
            if (responseCode == HttpURLConnection.HTTP_OK) { // success
                BufferedReader in = new BufferedReader(new InputStreamReader(
                        conn.getInputStream()));
                String inputLine;
                StringBuffer tempResponse = new StringBuffer();

                while ((inputLine = in.readLine()) != null) {
                    tempResponse.append(inputLine);
                }
                in.close();
                response = tempResponse.toString();
                System.out.println(response.toString());
            } else {
                System.out.println("GET request not worked");
            }
        } catch (IOException e) {
            e.printStackTrace();
        }

    }


}

答案 5 :(得分：0)

您必须提交请求。这可以通过调用getResponseCode（）或getResponseMessage（））或getInputStream（）来返回和处理响应来完成。

您的示例的工作代码：

import android.os.AsyncTask;
import android.os.Bundle;
import android.support.v7.app.AppCompatActivity;

import java.io.BufferedOutputStream;
import java.io.IOException;
import java.io.OutputStream;
import java.net.HttpURLConnection;
import java.net.ProtocolException;
import java.net.URL;

public class MainActivity extends AppCompatActivity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        toPost test = new toPost();
        test.execute();
    }

    private class toPost extends AsyncTask<URL, Void, String> {
        @Override
        protected String doInBackground(URL... params) {
            HttpURLConnection conn = null;
            try {
                URL url = new URL("http://example.com/service");
                conn = (HttpURLConnection) url.openConnection();
                conn.setReadTimeout(10000);
                conn.setConnectTimeout(15000);
                conn.setRequestMethod("POST");
                conn.setDoInput(true);
                conn.setDoOutput(true);
                String body = "username=user1&locationname=location1";
                OutputStream output = new BufferedOutputStream(conn.getOutputStream());
                output.write(body.getBytes());
                output.flush();

                //This is needed
                // Could alternatively use conn.getResponseMessage() or conn.getInputStream()
                conn.getResponseCode();

            } catch (ProtocolException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            } finally {
                conn.disconnect();
            }
            return null;
        }
    }
}

答案 6 :(得分：0)

试试这个。 （msg是xml字符串）

 try
      {
        URL url = new URL(address);
        URLConnection uc = url.openConnection();
        HttpURLConnection conn = (HttpURLConnection) uc;
        conn.setDoInput(true);
        conn.setDoOutput(true);
        conn.setRequestMethod("POST");
        conn.setRequestProperty("Content-type", "text/xml");        
        PrintWriter pw = new PrintWriter(conn.getOutputStream());
        pw.write(msg.getText());
        pw.close();
        BufferedInputStream bis = new BufferedInputStream(conn.getInputStream());
        bis.close();

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