如何在Java中解析XML根元素并选择它的一些值?
我正在寻找一种解析xml根元素并从中获取一些值的实用方法。 我试过很多方法,但没有一种是有效的如何在Java中解析XML根元素并选择它的一些值?,java,xml,jdom,Java,Xml,Jdom,我正在寻找一种解析xml根元素并从中获取一些值的实用方法。 我试过很多方法,但没有一种是有效的 DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance(); Document doc = factory.newDocumentBuilder().parse(fileLoaded); Element root = null;
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
Document doc = factory.newDocumentBuilder().parse(fileLoaded);
Element root = null;
NodeList list = doc.getChildNodes();
System.out.println(list.toString());
for (int i = 0; i < list.getLength(); i++) {
if (list.item(i) instanceof Element) {
root = (Element) list.item(i);
System.out.println(root.toString());
break;
}
}
root = doc.getDocumentElement();
}
DocumentBuilderFactory=DocumentBuilderFactory.newInstance();
Document doc=factory.newDocumentBuilder().parse(文件加载);
元素根=空;
NodeList list=doc.getChildNodes();
System.out.println(list.toString());
对于(int i=0;i
XML文件:
<planes_for_sale id="planeId" num="1452" name="boing">
<ad>
<year> 1977 </year>
<make> test </make>
<model> Skyhawk </model>
<color> Light blue and white </color>
<description> New paint, nearly new interior,
685 hours SMOH, full IFR King avionics </description>
<price> 23,495 </price>
<seller phone = "555-222-3333"> Skyway Aircraft </seller>
<location>
<city> Rapid City, </city>
<state> South Dakota </state>
</location>
1977
试验
天鹰
浅蓝色和白色
新油漆,新内饰,
685小时烟雾,全IFR国王航空电子设备
23,495
空中飞机
快速城市,
南达科他州
在我的例子中,我想从
planes\u for\u sale
根元素加载id
,num
,name
。尝试封送和解封机制
JAXBContext jaxbContext = JAXBContext.newInstance(EmployeeMap.class);
Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
EmployeeMap empMap = (EmployeeMap) jaxbUnmarshaller.unmarshal( new File("c:/temp/employees.xml") );
对于任何正在寻找实用方法的人来说,这里有一个测试:
DocumentBuilderFactory factory = DocumentBuilderFactory
.newInstance();
Document doc = factory.newDocumentBuilder().parse(fileLoaded);
doc.getDocumentElement().normalize();
System.out.println("Root element :"
+ doc.getDocumentElement().getNodeName());
NodeList nList = doc.getElementsByTagName("planes_for_sale");
System.out.println("----------------------------");
for (int temp = 0; temp < nList.getLength(); temp++) {
Node nNode = nList.item(temp);
System.out.println("\nCurrent Element :" + nNode.getNodeName());
if (nNode.getNodeType() == Node.ELEMENT_NODE) {
Element eElement = (Element) nNode;
System.out.println("Plane name : "
+ eElement.getAttribute("name"));
}
}
DocumentBuilderFactory=DocumentBuilderFactory
.newInstance();
Document doc=factory.newDocumentBuilder().parse(文件加载);
doc.getDocumentElement().normalize();
System.out.println(“根元素:”
+doc.getDocumentElement().getNodeName());
NodeList nList=doc.getElementsByTagName(“出售的飞机”);
System.out.println(“-------------------------------”;
对于(int-temp=0;temp
使用XPath提取属性值和元素内容
DocumentBuilder builder = DocumentBuilderFactory
.newInstance()
.newDocumentBuilder();
Document doc = builder.parse(...);
XPath xp = XPathFactory
.newInstance()
.newXPath();
String id = xp.evaluate("planes_for_sale/@id", doc);
String num = xp.evaluate("planes_for_sale/@num", doc);
String name = xp.evaluate("planes_for_sale/@name", doc);
System.out.println("id: " + id);
System.out.println("num: " + num);
System.out.println("name: " + name);
产生:
id: planeId
num: 1452
name: boing
看看这个答案
它向你展示了如何
- 将XML文件读入DOM
- 使用XPath筛选出一组节点
- 对每个提取的节点执行特定操作
thnx以获取您的答案,但在我的例子中,如果您看一下我的xml示例,它的xml结构就不一样了,因此我认为这样做是不对的。