Question

您好我需要解析xml并获取所有子节点以及节点<Employees>和</Employees>之间的数据

我有xml，如：

<?xml version="1.0"?>
<Employees>
    <Employee emplid="1111" type="admin">
        <firstname>test1</firstname>
        <lastname>Watson</lastname>
        <age>30</age>
        <email>johnwatson@sh.com</email>
    </Employee>
    <Employee emplid="2222" type="admin">
        <firstname>Sherlock</firstname>
        <lastname>Homes</lastname>
        <age>32</age>
        <email>sherlock@sh.com</email>
    </Employee>
</Employees>

我需要像

这样的回复

<Employee emplid="1111" type="admin">
        <firstname>test1</firstname>
        <lastname>Watson</lastname>
        <age>30</age>
        <email>johnwatson@sh.com</email>
    </Employee>
    <Employee emplid="2222" type="admin">
        <firstname>Sherlock</firstname>
        <lastname>Homes</lastname>
        <age>32</age>
        <email>sherlock@sh.com</email>
    </Employee>

我试过下面的代码

 FileInputStream file = new FileInputStream(new File("E:\\test.xml"));

         DocumentBuilderFactory builderFactory = DocumentBuilderFactory.newInstance();

         DocumentBuilder builder =  builderFactory.newDocumentBuilder();

         Document xmlDocument = builder.parse(file);

         XPath xPath =  XPathFactory.newInstance().newXPath();

         System.out.println("*************************");
         String expression = "/Employees/*";
         System.out.println(expression);
         String email = xPath.compile(expression).evaluate(xmlDocument);
         System.out.println(email);

但是我得到了回应

  test1
        Watson
        30
        johnwatson@sh.com

我使用了像 / Employees / * 这样的表达，但它不起作用

任何人都可以帮我这样做吗？

Answer 1

这将是一个可能的XSLT转换：

print("Total: ${}".format(total))

Answer 2

如果要将DOM节点序列化为字符串，请使用例如

import  org.w3c.dom.bootstrap.DOMImplementationRegistry;
import  org.w3c.dom.Document;
import  org.w3c.dom.ls.DOMImplementationLS;
import  org.w3c.dom.ls.LSSerializer;

...

DOMImplementationRegistry registry = DOMImplementationRegistry.newInstance();

DOMImplementationLS impl = 
    (DOMImplementationLS)registry.getDOMImplementation("LS");

LSSerializer writer = impl.createLSSerializer();
String str = writer.writeToString(node);

所以返回NodeList你可以使用

     String expression = "/Employees/*";
     System.out.println(expression);
     NodeList elements = (NodeList)xPath.compile(expression).evaluate(xmlDocument, XPathConstants.NODESET);
     for (int i = 0; i < elements.getLength(); i++) 
     {
       System.out.println(writer.writeToString(element.item(i));
     }

Answer 3

首先，如果您想匹配每个Employee，理想情况下，您的XPath表达式应该是Employee而不是/Employees/*。如果你知道标签名称，也不需要XPath，你可以xmlDocument.getElementsByTagName("Employee")。

如果要将节点序列化为String，可以使用Transformer，如下所示：

Transformer t = TransformerFactory.newTransformer();
NodeList nodes = xmlDocument.getElementsByTagName("Employee");
for(int i = 0; i < nodes.getLength(); i++) {
    StringWriter sw = new StringWriter();
    t.transform(new DOMSource(nodes.item(i)), new StreamResult(sw));
    String serialized = sw.toString();
    System.out.println(serialized);
}

解析XML以使用Java XPATH获取所有子节点及其数据

3 个答案: