使用android进行HTTP请求

我到处都找过了，但我找不到我的答案，有没有一种方法可以做一个简单的HTTP请求?我想在我的一个网站上请求一个PHP页面/脚本，但我不想显示网页。

如果可能的话，我甚至想在后台(在一个BroadcastReceiver)做它

当前回答

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这是一个非常古老的答案。我绝对不会再推荐Apache的客户端了。你可以用任意一种:

改造 OkHttp 截击 HttpUrlConnection

原来的答案

首先，申请访问网络的权限，在您的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Apache http客户端与Android捆绑:

    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，你可以通过以下方式提出请求:

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

2010-08-17 19:11:11

其他回答

我做了这个webservice请求URL，使用一个Gson库:

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;
}

2014-06-05 13:51:02

有一根线:

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);
    }
}

2010-08-17 19:03:32

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

2014-07-30 18:07:39

由于没有一个回答描述了一种使用OkHttp执行请求的方法，这是目前在Android和Java上非常流行的http客户端，我将提供一个简单的例子:

//get an instance of the client
OkHttpClient client = new OkHttpClient();

//add parameters
HttpUrl.Builder urlBuilder = HttpUrl.parse("https://www.example.com").newBuilder();
urlBuilder.addQueryParameter("query", "stack-overflow");


String url = urlBuilder.build().toString();

//build the request
Request request = new Request.Builder().url(url).build();

//execute
Response response = client.newCall(request).execute();

这个库的明显优势是，它将我们从一些低级细节中抽象出来，提供了更友好和安全的方式与它们交互。语法也被简化，允许编写漂亮的代码。

2018-06-11 18:59:43

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