Parsing XML on Android

2019-08-18 05:14发布

There are lots of tips to parse complicated XML but I want a parse for simple XML like:

<map>
     <string name="string_1">Hello there</string>
     <string name="string_2">This is Mium.</string>
</map>

I was trying to use SAXParser (I'm not sure which to use: SAXParser or XMLPullParser), so I'm looking at Parse XML on Android

public void ParseData(String xmlData)
{
    try
    {
        // Document Builder
        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
        DocumentBuilder db = factory.newDocumentBuilder();

        // Input Stream
        InputSource inStream = new InputSource();
        inStream.setCharacterStream(new StringReader(xmlData));

        // Parse Document into a NodeList
        Document doc = db.parse(inStream);
        NodeList nodes = doc.getElementsByTagName("ticket");

        // Loop NodeList and Retrieve Element Data
        for(int i = 0; i < nodes.getLength(); i++)
        {
            Node node = nodes.item(i);

            if (node instanceof Element)
            {
                Element child = (Element)node;
                String id = child.getAttribute("id");
            }
        }
    }
    catch(SAXException e)
    {

    }
    }

However, it's already complicated XML parse. How should I code only to parse ?

1条回答
淡お忘
2楼-- · 2019-08-18 06:01
Your loop should look like this:

Node node = nList.item(i);
if (node.getNodeType() == Node.ELEMENT_NODE) 
{
     Element element = (Element) node;
     tv1.setText(tv1.getText()+"\nName : " + getValue("name", element)+"\n");
}

where getValue :

private static String getValue(String tag, Element element)
{
   NodeList nodeList = element.getElementsByTagName(tag).item(0).getChildNodes();
   Node node = nodeList.item(0);
   return node.getNodeValue();
}
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