更新

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");

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。

这是android中HTTP Get/POST请求的新代码。HTTPClient已被废弃，可能无法使用，因为在我的情况下。

首先在build.gradle中添加两个依赖:

compile 'org.apache.httpcomponents:httpcore:4.4.1'
compile 'org.apache.httpcomponents:httpclient:4.5'

然后在ASyncTask in doBackground方法中编写此代码。

 URL url = new URL("http://localhost:8080/web/get?key=value");
 HttpURLConnection urlConnection = (HttpURLConnection)url.openConnection();
 urlConnection.setRequestMethod("GET");
 int statusCode = urlConnection.getResponseCode();
 if (statusCode ==  200) {
      InputStream it = new BufferedInputStream(urlConnection.getInputStream());
      InputStreamReader read = new InputStreamReader(it);
      BufferedReader buff = new BufferedReader(read);
      StringBuilder dta = new StringBuilder();
      String chunks ;
      while((chunks = buff.readLine()) != null)
      {
         dta.append(chunks);
      }
 }
 else
 {
     //Handle else
 }

注意:与Android捆绑的Apache HTTP客户端现在已弃用，转而支持HttpURLConnection。请参阅Android开发者博客了解更多细节。

添加<uses-permission android:name="android.permission. "INTERNET" />到您的舱单。

然后你会像这样检索一个网页:

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请求的更多信息，请参阅文档。

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

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;
}