怎么获取这个xml标签里的值啊

如何获取这个xml标签里的值啊？
有工程1. grpIndustryWeb_1 Location为E:\program\grpSmartEnterpriseApp\WebContent\WEB-INF\web.xml
  2. z_mytest Location为E:\program\z_mytest\src\Test1.java


/grpSmartEnterpriseApp/WebContent/WEB-INF/web.xml


我想在/z_mytest/src/Test1.java里面写代码，
其功能是读取/grpSmartEnterpriseApp/WebContent/WEB-INF/web.xml里面的两个标签。

请教怎么写啊？

Java code

public class Test1 {    public static void main(String args[]) {        //这里先找到文件web.xml    /*    读下面这个标签：        <display-name>grpSmartEnterpriseApp</display-name>    */    String displayName=//这里获取上面标签里的值：grpSmartEnterpriseApp    /*    读下面这个标签：    <servlet-mapping>        <servlet-name>Schema_1Service</servlet-name>        <url-pattern>/servlet/schema_1/Service</url-pattern>    </servlet-mapping>    */    String urlPattern=//这里获取上面标签里的值：/servlet/schema_1/Service        String myurl="http://localhost:8080/";    myurl=myrul+servletName+urlPattern;    System.out.println(myurl);        }}

/grpSmartEnterpriseApp/WebContent/WEB-INF/web.xml内容如下：

<?xml version="1.0" encoding="UTF-8"?>

XML code

<web-app id="web-app_1" xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"    xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd" version="2.4">    <display-name>grpSmartEnterpriseApp</display-name>    <distributable />        <listener>        <listener-class>grp.servlet.GrpSessionListener</listener-class>    </listener>    <servlet>        <display-name>Schema_1CatalogueServlet</display-name>        <servlet-name>Schema_1CatalogueServlet</servlet-name>        <servlet-class>grp.industry.schema_1.servlets.CatalogueServlet</servlet-class>    </servlet>    <!--省略部分标签-->        <servlet-mapping>        <servlet-name>Schema_1Service</servlet-name>        <url-pattern>/servlet/schema_1/Service</url-pattern>    </servlet-mapping>    <servlet-mapping>        <servlet-name>LoginServlet</servlet-name>        <url-pattern>/servlet/security/Login</url-pattern>    </servlet-mapping></web-app>

[解决办法]

Java code

        try        {            SAXReader reader = new SAXReader();            Document document = reader.read(new File("F://web.xml")); //这里记着把文件路径改了                        Element root = document.getRootElement();            Element dnE = root.element("display-name");            String displayName = dnE.getText();                        Element smE = root.element("servlet-mapping");            Element upE = smE.element("url-pattern");            String urlPattern = upE.getText();                        System.out.println(displayName);            System.out.println(urlPattern);                    }        catch(DocumentException e)        {            e.printStackTrace();        }
[解决办法]
HTML code
<?xml version="1.0" encoding="UTF-8" ?><list> <serverGroup name="default">   <server hostname ="192.168.0.0" ip="192.168.0.0" port="1024"/>  </serverGroup>  <serverGroup name="servergroup1">   <server hostname ="192.168.0.1" ip="192.168.0.1" port="1024"/>   <server hostname ="192.168.0.2" ip="192.168.0.2" port="1024"/>   <server hostname ="192.168.0.3" ip="192.168.0.3" port="1024"/>  </serverGroup>  <serverGroup name="servergroup1">   <server hostname ="192.168.0.4" ip="192.168.0.4" port="1024"/>   <server hostname ="192.168.0.5" ip="192.168.0.5" port="1024"/>   </serverGroup></list> 
 
[解决办法]
XML 解析建议LZ可以到http://www.w3school.com.cn/xmldom/index.asp看一下 . 

 

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