Как легче всего распарсить xml документ?

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12

import xml.etree.ElementTree as ET

# Парсинг XML
tree = ET.parse('file.xml')
root = tree.getroot()

# Обращение к элементам XML
for child in root:
    print(child.tag, child.attrib)

# Получение данных из XML
data = root.find('tag_name').text

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34

import java.io.File;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import org.w3c.dom.Document;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;

public class XMLParser {

    public static void main(String[] args) {
        try {
            File file = new File("file.xml");
            DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
            DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
            Document doc = dBuilder.parse(file);

            doc.getDocumentElement().normalize();

            System.out.println("Root element: " + doc.getDocumentElement().getNodeName());

            NodeList nList = doc.getElementsByTagName("tag_name");

            for (int i = 0; i < nList.getLength(); i++) {
                Node node = nList.item(i);
                if (node.getNodeType() == Node.ELEMENT_NODE) {
                    System.out.println("Tag name: " + node.getNodeName());
                    System.out.println("Tag value: " + node.getTextContent());
                }
            }
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}