Question

我是XML Parsing的新手，我需要获取url feed并解析它们，然后将它们保存在本地数据库中。如果我传递一些其他链接它接受，但如果我通过我需要的链接它给出这个错误。我还提供了我的代码部分和log cat输出。谁能告诉我我面临的问题是什么？

以下是我需要解析的网址。

private static final String POSTS_URL = "http://bgr.in/feed/iphoneapp";

这是产生错误的方法： -

public static void updateStories(Context context, String url, int type) {
    if (StoryService.isConnected(context)) {
        HttpGet request = new HttpGet(url);
        request.addHeader("accepts", "application/rss+xml");
        HttpClient client = new DefaultHttpClient();

        HttpResponse response = null;
        try {
            response = client.execute(request);
        } catch (IOException e) {
            Log.e(TAG, "Failed to get rss feed.", e);
        }

        RootElement root = new RootElement("rss");

        List<BaseStory> stories = Story.appendArrayListener(type, 
                root.getChild("channel"), 0);

        try {
            Xml.parse(response.getEntity().getContent(), Xml.Encoding.UTF_8, root.getContentHandler());
        } catch (Exception e) {
            e.printStackTrace();
            Log.i(TAG, "Failed to parse " + url, e);
        }`public static void updateStories(Context context, String url, int type) {
    if (StoryService.isConnected(context)) {
        HttpGet request = new HttpGet(url);
        request.addHeader("accepts", "application/rss+xml");
        HttpClient client = new DefaultHttpClient();

        HttpResponse response = null;
        try {
            response = client.execute(request);
        } catch (IOException e) {
            Log.e(TAG, "Failed to get rss feed.", e);
        }

        RootElement root = new RootElement("rss");

        List<BaseStory> stories = Story.appendArrayListener(type, 
                root.getChild("channel"), 0);

        try {
            Xml.parse(response.getEntity().getContent(), Xml.Encoding.UTF_8, root.getContentHandler());
        } catch (Exception e) {
            e.printStackTrace();
            Log.i(TAG, "Failed to parse " + url, e);
        }

Log Cat输出如下： -

System.err(19251):  org.apache.harmony.xml.ExpatParser$ParseException: At line 292,column 0: junk after document element
System.err(19251):  at org.apache.harmony.xml.ExpatParser.parseFragment(ExpatParser.java:515)
System.err(19251):  at org.apache.harmony.xml.ExpatParser.parseDocument(ExpatParser.java:474)
System.err(19251):  at org.apache.harmony.xml.ExpatReader.parse(ExpatReader.java:316)
System.err(19251):  at org.apache.harmony.xml.ExpatReader.parse(ExpatReader.java:279)
System.err(19251):  at android.util.Xml.parse(Xml.java:84)
System.err(19251):  at in.bgr.service.StoryService.updateStories(StoryService.java:122)
System.err(19251):  at in.bgr.service.StoryService.updateStories(StoryService.java:96)
System.err(19251):  at in.bgr.ui.AllStoriesFragment$LoadStoriesTask.doInBackground(AllStoriesFragment.java:309)
System.err(19251):  at in.bgr.ui.AllStoriesFragment$LoadStoriesTask.doInBackground(AllStoriesFragment.java:1)
System.err(19251):  at android.os.AsyncTask$2.call(AsyncTask.java:288)
System.err(19251):  at java.util.concurrent.FutureTask.run(FutureTask.java:237)
System.err(19251):  at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:231)
System.err(19251):  at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1112)

Answer 1

“文档元素之后的垃圾”错误是在根元素标记之后的xml中有任何不需要的标记时引起的。只删除根元素标记后面的所有标记。在我的例子中，XML元素在XML文件中的行号292处重复输入。我清除了它，现在我觉得它对我有用。

org.apache.harmony.xml.ExpatParser $ ParseException：在第292行，第0列：文档元素之后的垃圾

1 个答案: