Question

我正在开发一个Android应用程序，我想在没有打开android的默认浏览器的情况下从网址获取新闻。这意味着我只想获取texutal内容，只获取新闻而不是浏览器中的完整html页面。我怎么能这样做？

Answer 1

如果我理解正确 - 您需要在线提出请求并收到html代码。这样做如下：

DefaultHttpClient client = new DefaultHttpClient();
HttpGet request = new HttpGet();
request.setURI(new URI(http://example.com/news));

HttpResponse response = client.execute(request);
BufferedReader in = new BufferedReader(new InputStreamReader(
response.getEntity().getContent()));
StringBuffer sb = new StringBuffer("");
String line;
while ((line = in.readLine()) != null) {
    sb.append(line);
}
in.close();

String html = sb.toString();

Answer 2

您是否想要将网页的实际内容解析为您的应用程序？当我在我的一个应用程序中这样做时，我用一个简单的http://developer.android.com/reference/org/xmlpull/v1/XmlPullParser.html解析了整个网页，然后我拿出那些相关的标签。然而，这需要在asynctask（http://developer.android.com/reference/android/os/AsyncTask.html）下运行一些非常繁重的动态编程。

URL url = new URL(XML_INIT_ADRESS);
XmlPullParserFactory factory = XmlPullParserFactory.newInstance();
factory.setNamespaceAware(false);
XmlPullParser xpp = factory.newPullParser();
xpp.setInput(url.openStream(), null);

我个人对Android还没有很熟悉，我还在学习，但你应该能够通过这种方式解析网页上的新闻。

编辑：这种方法几乎需要对某些“新闻标签”进行某种识别，如果它们“无法确定”，安东斯的答案会更好。

Answer 3

    Hi Yes You can implement this.Use This code which i mention below.



 - WebView webView = (WebView) findViewById(R.id.webview_btn);
   WebSettings settings = webView.getSettings();
   mProgress = ProgressDialog.show(this, "Loading", "Please wait for a moment...");

               settings.setJavaScriptEnabled(true);
               settings.setSupportZoom(true);
               settings.setBuiltInZoomControls(true);

               WebViewClient client = new WebViewClient() {
                @Override
                public boolean shouldOverrideUrlLoading(WebView view, String url) {
                    // TODO Auto-generated method stub
                    view.loadUrl(url);
                    mProgress.show();

                    return true;
                }
                   @Override
                   public void onPageFinished(WebView view, String url) {

                    if(mProgress.isShowing())
                    {
                        mProgress.dismiss();
                    }
                   }
               };
               webView.loadUrl(enter your url);
               webView.setWebViewClient(client);

Answer 4

HttpClient httpClient = new DefaultHttpClient();
HttpContext localContext = new BasicHttpContext();
HttpGet httpGet = new HttpGet("http://www.example.com/" + URL);
response = httpClient.execute(httpGet, localContext);

在没有安卓android浏览器帮助的情况下打开网址？

4 个答案: