我正在构建一个狩猎应用程序,我需要以this站点的RSS源形式提供天气信息。
我使用了this网站上的代码并列出了Feed,但是当我点击某个项目时,它并没有将其连接到该网站以获取更多信息。我想获得温度和风信息,但我不知道如何,因为我是编程的初学者。
我非常感谢任何帮助,特别是以解决我的问题的代码形式。
答案 0 :(得分:4)
以下是代码从URL获取RSS Feed并将其列在android中的listview中。
首先需要创建一个扩展ListActivity的类,然后输入以下代码:
// Initializing instance variables
headlines = new ArrayList();
links = new ArrayList();
try {
URL url = new URL("http://www.RSS-Feed-URL-HERE");
XmlPullParserFactory factory = XmlPullParserFactory.newInstance();
factory.setNamespaceAware(false);
XmlPullParser xpp = factory.newPullParser();
// We will get the XML from an input stream
xpp.setInput(getInputStream(url), "UTF_8");
/* We will parse the XML content looking for the "<title>" tag which appears inside the "<item>" tag.
* However, we should take in consideration that the rss feed name also is enclosed in a "<title>" tag.
* As we know, every feed begins with these lines: "<channel><title>Feed_Name</title>...."
* so we should skip the "<title>" tag which is a child of "<channel>" tag,
* and take in consideration only "<title>" tag which is a child of "<item>"
*
* In order to achieve this, we will make use of a boolean variable.
*/
boolean insideItem = false;
// Returns the type of current event: START_TAG, END_TAG, etc..
int eventType = xpp.getEventType();
while (eventType != XmlPullParser.END_DOCUMENT) {
if (eventType == XmlPullParser.START_TAG) {
if (xpp.getName().equalsIgnoreCase("item")) {
insideItem = true;
} else if (xpp.getName().equalsIgnoreCase("title")) {
if (insideItem)
headlines.add(xpp.nextText()); //extract the headline
} else if (xpp.getName().equalsIgnoreCase("link")) {
if (insideItem)
links.add(xpp.nextText()); //extract the link of article
}
}else if(eventType==XmlPullParser.END_TAG && xpp.getName().equalsIgnoreCase("item")){
insideItem=false;
}
eventType = xpp.next(); //move to next element
}
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (XmlPullParserException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
// Binding data
ArrayAdapter adapter = new ArrayAdapter(this,
android.R.layout.simple_list_item_1, headlines);
setListAdapter(adapter);
}
public InputStream getInputStream(URL url) {
try {
return url.openConnection().getInputStream();
} catch (IOException e) {
return null;
}
}
@Override
protected void onListItemClick(ListView l, View v, int position, long id) {
Uri uri = Uri.parse((String) links.get(position));
Intent intent = new Intent(Intent.ACTION_VIEW, uri);
startActivity(intent);
}
答案 1 :(得分:1)
您可以在google play下载并检查我的项目。这个项目是关于一些土耳其体育频道。它就像所有应用程序一样。
您可以在github上查看项目的源代码。
答案 2 :(得分:-1)
您可以访问任何热门的天气网站,例如Weather.com,并查找其API部分。大多数网站都提供用于开发应用程序的API。 Weather.com的API部分实际上指导您完成实施过程。