XML格式
<?xml version="1.0"?>
<Employees>
<Employee emplid="1111" type="admin">
<firstname>John</firstname>
<lastname>Watson</lastname>
<age>30</age>
<email>[email protected]</email>
</Employee>
<Employee emplid="2222" type="admin">
<firstname>Sherlock</firstname>
<lastname>Homes</lastname>
<age>32</age>
<email>[email protected]</email>
</Employee>
<Employee emplid="3333" type="user">
<firstname>Jim</firstname>
<lastname>Moriarty</lastname>
<age>52</age>
<email>[email protected]</email>
</Employee>
<Employee emplid="4444" type="user">
<firstname>Mycroft</firstname>
<lastname>Holmes</lastname>
<age>41</age>
<email>[email protected]</email>
</Employee>
</Employees>
代码
FileInputStream file = new FileInputStream(new File("/Users/Desktop/employees.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/Employee/firstname";
System.out.println(expression);
NodeList nodeList = (NodeList) xPath.compile(expression).evaluate(xmlDocument, XPathConstants.NODESET);
for (int i = 0; i < nodeList.getLength(); i++) {
System.out.println(nodeList.item(i).getFirstChild().getNodeValue());
}
通过使用上面的代码，我设法获取了John,Sherlock,Jim,Mycroft。如果我想得到我该怎么办emplid="1111" type="admin" John,emplid="2222" type="admin" Sherlock,emplid="3333" type="user" Jim,emplid="4444" type="user" Mycroft。任何建议或参考链接都将受到高度赞赏。
看看@ API：
String expression = "/Employees/Employee";
NodeList nodeList = (NodeList) xPath.compile(expression).evaluate(xmlDocument, XPathConstants.NODESET);
for (int i = 0; i < nodeList.getLength(); i++) {
Node employeeNode = nodeList.item(i);
String emplId = employeeNode.getAttributes().getNamedItem("emplid").getNodeValue();
}
（http://docs.oracle.com/javase/6/docs/api/org/w3c/dom/Node.html#getAttributes（））
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