1

我在 android 中使用 SAX 解析器从 url 获取 xml 数据，但我仍然没有得到响应的存储位置。

代码：

String c=city.getText().toString();
String s=state.getText().toString();
StringBuilder URL=new StringBuilder(baseURL);
URL.append(c+","+s);
try{
    SAXParserFactory factory=SAXParserFactory.newInstance();
    SAXParser parser=factory.newSAXParser();
    XMLReader reader=parser.getXMLReader();
    HandlingXMLStuff handler=new HandlingXMLStuff();
    reader.setContentHandler(handler);
    URL url = new
    URL("http://api.openweathermap.org/data/2.5/weathermode=xml&q="+c+","+s);
    HttpURLConnection connection =  (HttpURLConnection)url.openConnection();
    reader.parse(new InputSource(connection.getInputStream()));
String information=handler.getInformation();
output.setText(information);
}
catch(Exception e)
{
    output.setText(e.toString());
}

我的问题是响应存储在哪里以及如何访问它以从 xml 获取所需的信息？

编辑：非常感谢大家的帮助..现在我理解了逻辑..但我面临另一个问题....

代码：

String c = city.getText().toString();
    String s = state.getText().toString();
    try {
        SAXParserFactory factory = SAXParserFactory.newInstance();
        SAXParser parser = factory.newSAXParser();
        XMLReader reader = parser.getXMLReader();
        HandlingXMLStuff handler = new HandlingXMLStuff();
        reader.setContentHandler(handler);
        String query = URLEncoder.encode(
        "http://api.openweathermap.org/data/2.5/weather?mode=xml&q="
                        + c + "," + s, "utf-8");
URL url = new URL("http://api.openweathermap.org/data/2.5/weather?mode=xml&q="
                + c + "," + s);
        HttpURLConnection connection = (HttpURLConnection) url
                .openConnection();
        reader.parse(new InputSource(connection.getInputStream()));
        String information = handler.getInformation();
        output.setText(information);
    } catch (Exception e) {
        output.setText(e.toString());
    }

我没有收到来自 url 的响应，出现错误“无法将 null 解析为 int”。知道发生了什么吗？我在浏览器中输入了相同的 url 并得到了正确的响应，但在这里不起作用。

编辑：固定。我的处理程序函数 xD 出现一些错误

4

1 回答 1

0 
feedParser oFeedParser = new feedParser(MainActivity.this);
                try {
                    /**
                    * Create a new instance of the SAX parser
                    **/
                    SAXParserFactory saxPF = SAXParserFactory.newInstance();
                    SAXParser saxP = saxPF.newSAXParser();
                    XMLReader xmlR = saxP.getXMLReader();
                    URL url = new URL("Your_URL"); // URL of the XML

                    xmlR.setContentHandler(oFeedParser);
                    xmlR.parse(new InputSource(url.openStream()));
                    handler.sendEmptyMessage(0);

                } catch(IOException e) {
                    System.out.println(e);
                    handler.sendEmptyMessage(1);
                } catch (Exception e) {
                    System.out.println(e);
                } finally {
                    oFeedParser.closeDataBase();
                }
            }

和 feedParser 类

public class feedParser extends DefaultHandler {

    String elementValue = null;
    boolean elementOn, isFeedInitialized = false;
    feed oFeed;
    feedsDAO dao;

    public feedParser(Context context) {
        dao = new feedsDAO(context);
        dao.open();
    }

    public void closeDataBase() {
        dao.close();
    }

    /**
     * This will be called when the tags of the XML starts.
     **/
    @Override
    public void startElement(String uri, String localName, String qName, Attributes attributes) throws SAXException {

        if (localName.equals("ITEM_NAME")) {
        } 
    }

    /**
     * This will be called when the tags of the XML end.
     **/
    @Override
    public void endElement(String uri, String localName, String qName) throws SAXException {

        if (localName.equals("ITEM_NAME")) {

        }
    }

    /**
     * This is called to get the tags value
     **/
    @Override
    public void characters(char[] ch, int start, int length)
            throws SAXException {
        if (elementOn) {
            elementValue = new String(ch, start, length);
            elementOn = false;
        }
    }
}
于 2013-10-01T11:13:44.380 回答