Question

我希望我的Android应用程序发出以下/共享请求（POST）。 https://www.dropbox.com/developers/reference/api#shares

但我以前没有做任何http请求，也不知道如何。

我的应用已经使用Dropbox进行了身份验证。

有人可以提供样品吗？

ps.i知道http的理论但不是它在java中的实际用途

Answer 1

我的建议是使用像LoopJ这样的库。它将处理你不想自己实现的事情，如“请求重试”。它附带了本页面上的简单示例。

http://loopj.com/android-async-http/

Answer 2

Http请求在Android中以这种方式完成，它只是一个示例代码，您尝试了很多相关的事情。

有用的指南：http://developer.android.com/reference/org/apache/http/client/HttpClient.html

        HttpClient httpclient = new DefaultHttpClient();   
        HttpPost httpPost = new HttpPost(YOUR_URL);
        HttpResponse response;
        try {
            response = httpclient.execute(httpPost); // the request executes
            Log.d("HTTP","Executed");
    String  responseBody = EntityUtils.toString(response.getEntity());

        } catch (ClientProtocolException e) {
            e.printStackTrace();
        }
        catch(ConnectTimeoutException e){
        e.printStackTrace();
        }
        catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }

希望这有助于

Answer 3

您使用以下示例。

此示例用于读取http web服务中的json字符串

public class Httprequest_responseActivity extends Activity {
ProgressDialog progressdialog;
TextView txt;
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    txt=(TextView)findViewById(R.id.txt);
    httprequest("http://api.bilubi.com/BBService.svc/Compleateprogress");
}
String urlstr;
public void httprequest(String url)
{
    urlstr=url;
    progressdialog=ProgressDialog.show(Httprequest_responseActivity.this, "", "Loadding........",true);
    new Thread(new Runnable() {

        @Override
        public void run() {
            BufferedReader in=null;

            Message msg=Message.obtain();
            msg.what=1;
            try
            {
                HttpClient client=new DefaultHttpClient();
                HttpGet request=new HttpGet();
                request.setURI(new URI(urlstr));
                HttpResponse response=client.execute(request);
                in=new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
                StringBuffer sb=new StringBuffer("");
                String line="";
                while((line=in.readLine())!=null)
                    sb.append(line);
                Bundle b=new Bundle();
                b.putString("data", sb.toString());
                msg.setData(b);
                in.close();

            }
            catch(Exception e)
            {
                System.out.println("****************"+e.getMessage());
                //txt.setText(""+e.getMessage());
            }

            handler.sendMessage(msg);
        }
    }).start(); 
}

Handler handler=new Handler()
{
    public void handleMessage(Message msg)
    {
        super.handleMessage(msg);
        switch(msg.what)
        {
        case 1:

            txt.setText(msg.getData().getString("data"));
            break;
        }
        progressdialog.dismiss();
    }
};

}

Answer 4

使用HttpURLConnection而不是HttpClient，这是Android开发人员推荐的： http://android-developers.blogspot.in/2011/09/androids-http-clients.html

以下是样本：

URL url = new URL("www.yandex.ru");
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
InputStream in = new BufferedInputStream(connection.getInputStream());
String response = new Scanner(in).useDelimiter("\\A").next();

我在哪里可以学习在Android中发出http请求？

4 个答案: