Question

我在资源/ raw上使用xml文档以填充ListView，这是代码：

List<Item> items = new ArrayList<Item>();
    try {
        BufferedReader br = new BufferedReader(new InputStreamReader(getResources().openRawResource(R.raw.data)));
        XmlPullParser p = XmlPullParserFactory.newInstance().newPullParser();
        p.setInput(br);
        int tipoevento = p.next();
        while (tipoevento != XmlPullParser.END_DOCUMENT) {
            switch (tipoevento) {
                case XmlPullParser.START_TAG:
                    if (p.getAttributeValue(null,"id") != null) {
                        items.add(
                                new Item(
                                        Integer.parseInt(p.getAttributeValue(null, "id")),
                                        p.getAttributeValue(null, "name"),
                                        p.getAttributeValue(null, "url")));
                    }
                    break;
                default:break;
            }
            tipoevento=p.next();
        }
        br.close();
    }
    catch(Exception e) {
        Log.e("Error: ", e.getMessage());}

而且，当我运行它时，我收到以下错误：Error :: Invalid int：“R.drawable.following”。我以这种方式将资源id放在xml中，我知道错误在解析中，但我不知道该怎么做：

<?xml version="1.0" encoding="utf-8"?>
<filmList>
<film name="Following" id="R.drawable.following" url="http://www.imdb.com/title/tt0154506/"/>
<film name="Memento" id="R.drawable.memento" url="http://www.imdb.com/title/tt0209144/"/>
<film name="Batman Begins" id="R.drawable.batman_begins" url="http://www.imdb.com/title/tt0372784/"/>
<film name="The Prestige" id="R.drawable.the_prestige" url="http://www.imdb.com/title/tt0482571/"/>
<film name="The Dark Knight" id="R.drawable.the_dark_knight" url="http://www.imdb.com/title/tt0468569/"/>
<film name="Inception" id="R.drawable.inception" url="http://www.imdb.com/title/tt1375666/"/>
<film name="The Dark Knight Rises" id="R.drawable.the_dark_knight_rises" url="http://www.imdb.com/title/tt1345836/"/>
</filmList>

从xml文档中读取的可绘制资源

0 个答案: