用android发出HTTP请求

我到处搜索但我找不到答案，有没有办法发出简单的HTTP请求？ 我想在我的某个网站上请求PHP页面/脚本，但我不想显示该网页。

如果可能的话我甚至想在后台（在BroadcastReceiver中）这样做

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#1楼

我使用Gson lib为web服务创建了这个URL：

客户：

public EstabelecimentoList getListaEstabelecimentoPorPromocao(){

        EstabelecimentoList estabelecimentoList  = new EstabelecimentoList();
        try{
            URL url = new URL("http://" +  Conexao.getSERVIDOR()+ "/cardapio.online/rest/recursos/busca_estabelecimento_promocao_android");
            HttpURLConnection con = (HttpURLConnection) url.openConnection();

            if (con.getResponseCode() != 200) {
                    throw new RuntimeException("HTTP error code : "+ con.getResponseCode());
            }

            BufferedReader br = new BufferedReader(new InputStreamReader((con.getInputStream())));
            estabelecimentoList = new Gson().fromJson(br, EstabelecimentoList.class);
            con.disconnect();

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

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#2楼

private String getToServer(String service) throws IOException {
    HttpGet httpget = new HttpGet(service);
    ResponseHandler<String> responseHandler = new BasicResponseHandler();
    return new DefaultHttpClient().execute(httpget, responseHandler);

}

问候

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#3楼

注意：现在不推荐使用与Android捆绑在一起的Apache HTTP Client，而使用HttpURLConnection 。 有关详细信息，请参阅Android开发人员博客 。

将<uses-permission android:name="android.permission.INTERNET" />到清单中。

然后，您将检索如下所示的网页：

URL url = new URL("http://www.android.com/");
HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
try {
     InputStream in = new BufferedInputStream(urlConnection.getInputStream());
     readStream(in);
}
finally {
     urlConnection.disconnect();
}

我还建议在一个单独的线程上运行它：

class RequestTask extends AsyncTask<String, String, String>{

@Override
protected String doInBackground(String... uri) {
    String responseString = null;
    try {
        URL url = new URL(myurl);
        HttpURLConnection conn = (HttpURLConnection) url.openConnection();
        if(conn.getResponseCode() == HttpsURLConnection.HTTP_OK){
            // Do normal input or output stream reading
        }
        else {
            response = "FAILED"; // See documentation for more info on response handling
        }
    } catch (ClientProtocolException e) {
        //TODO Handle problems..
    } catch (IOException e) {
        //TODO Handle problems..
    }
    return responseString;
}

@Override
protected void onPostExecute(String result) {
    super.onPostExecute(result);
    //Do anything with response..
}
}

有关响应处理和POST请求的详细信息，请参阅文档 。

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#4楼

有一个线程：

private class LoadingThread extends Thread {
    Handler handler;

    LoadingThread(Handler h) {
        handler = h;
    }
    @Override
    public void run() {
        Message m = handler.obtainMessage();
        try {
            BufferedReader in = 
                new BufferedReader(new InputStreamReader(url.openStream()));
            String page = "";
            String inLine;

            while ((inLine = in.readLine()) != null) {
                page += inLine;
            }

            in.close();
            Bundle b = new Bundle();
            b.putString("result", page);
            m.setData(b);
        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

        handler.sendMessage(m);
    }
}

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#5楼

UPDATE

这是一个非常古老的答案。 我绝对不会再推荐Apache的客户了。 而是使用：

• OkHttp
• HttpURLConnection的

原始答案

首先，请求访问网络的权限，在清单中添加以下内容：

<uses-permission android:name="android.permission.INTERNET" />

那么最简单的方法是使用与Android捆绑的Apache http客户端：

    HttpClient httpclient = new DefaultHttpClient();
    HttpResponse response = httpclient.execute(new HttpGet(URL));
    StatusLine statusLine = response.getStatusLine();
    if(statusLine.getStatusCode() == HttpStatus.SC_OK){
        ByteArrayOutputStream out = new ByteArrayOutputStream();
        response.getEntity().writeTo(out);
        String responseString = out.toString();
        out.close();
        //..more logic
    } else{
        //Closes the connection.
        response.getEntity().getContent().close();
        throw new IOException(statusLine.getReasonPhrase());
    }

如果你想让它在单独的线程上运行，我建议扩展AsyncTask：

class RequestTask extends AsyncTask<String, String, String>{

    @Override
    protected String doInBackground(String... uri) {
        HttpClient httpclient = new DefaultHttpClient();
        HttpResponse response;
        String responseString = null;
        try {
            response = httpclient.execute(new HttpGet(uri[0]));
            StatusLine statusLine = response.getStatusLine();
            if(statusLine.getStatusCode() == HttpStatus.SC_OK){
                ByteArrayOutputStream out = new ByteArrayOutputStream();
                response.getEntity().writeTo(out);
                responseString = out.toString();
                out.close();
            } else{
                //Closes the connection.
                response.getEntity().getContent().close();
                throw new IOException(statusLine.getReasonPhrase());
            }
        } catch (ClientProtocolException e) {
            //TODO Handle problems..
        } catch (IOException e) {
            //TODO Handle problems..
        }
        return responseString;
    }

    @Override
    protected void onPostExecute(String result) {
        super.onPostExecute(result);
        //Do anything with response..
    }
}

然后您可以通过以下方式提出请求：

   new RequestTask().execute("http://stackoverflow.com");

来源：oschina

链接：https://my.oschina.net/u/3797416/blog/3185317