Question

我写了下面这段代码。从互联网上找到它。这是我第一次使用XML Parsers。

public class MainActivity extends ListActivity {

List headlines;
List links;

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    headlines = new ArrayList();
    links = new ArrayList();
    try {
        URL url = new URL("http://www.sportlarissa.gr/feed/");

        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");

        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);
}

private 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);
}
}

xml文件如下。我很讨厌我在那里做错了什么，并且没有看到它!!

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="fill_parent"
android:layout_height="fill_parent"
android:orientation="vertical" >

<ListView
android:id="@+android:id/list"
android:layout_width="fill_parent"
android:layout_height="wrap_content" >
</ListView>

</LinearLayout>

Logcat输出：

06-13 08:01:09.147: E/AndroidRuntime(1392): FATAL EXCEPTION: main
06-13 08:01:09.147: E/AndroidRuntime(1392): java.lang.RuntimeException: Unable to start
activity 
ComponentInfo{com.example.rssfeederproject/com.example.rssfeederproject.MainActivity}:
java.lang.RuntimeException: Your content must have a ListView whose id attribute is
'android.R.id.list'

我有ListView，声明为

android:id="@+android:id/list"

为什么会这样？非常感谢提前:)）

Answer 1

删除+：必须将ID声明为"@android:id/list"。

Rss馈线示例给出例外

1 个答案: