使用JAXP(JEE6)解析一个简单的XML文档

我想为我的网络应用程序创建一个授权过滤器(以便能够限制对某些页面的访问).

我创建了一个简单的.xml文件,其中包含允许每个用户访问的页面：

<access>
    <buyer>
        <page>buyoffer.xhtml</page>
        <page>faq.xhtml</page>
        <page>index.jsp</page>
        <page>login.xhtml</page>
        <page>main.xhtml</page>
        <page>registrationSucceded.xhtml</page>     
    </buyer>
    <seller>
        <page>sellerpanel.xhtml</page>
        <page>faq.xhtml</page>
        <page>index.jsp</page>
        <page>login.xhtml</page>
        <page>main.xhtml</page>
        <page>registrationSucceded.xhtml</page>     
    </seller>
    <administrator>
        <page>sellerpanel.xhtml</page>
        <page>faq.xhtml</page>
        <page>index.jsp</page>
        <page>login.xhtml</page>
        <page>main.xhtml</page>
        <page>registrationSucceded.xhtml</page>     
    </administrator>
</access>

然后我需要进行解析以提取页面的值,以便能够创建允许或重定向的条件(依赖).我只需要告诉某人如何从xml中提取这些页面的值.这就是我现在所做的：

public class RestrictPageFilter implements Filter {

    private FilterConfig fc;
    private DocumentBuilder builder;
    private Document document;

    public void init(FilterConfig filterConfig) throws ServletException {
        // The easiest way to initialize the filter
        fc = filterConfig;
        // Get the file that contains the allowed pages
        File f = new File("/allowedpages.xml");
        // Prepare the file parsing
        try {
            builder = DocumentBuilderFactory.newInstance().newDocumentBuilder();
            document = builder.parse(f);
        } catch (ParserConfigurationException e) {
            e.printStackTrace();
        } catch (SAXException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

    }

    public void doFilter(ServletRequest request,ServletResponse response,FilterChain chain) throws IOException,ServletException {

        HttpServletRequest req = (HttpServletRequest) request;
        HttpServletResponse resp = (HttpServletResponse) response;
        HttpSession session = req.getSession(true);
        String pageRequested = req.getRequestURL().toString();

        // Get the value of the current logged user
        Role currentUser = (Role) session.getAttribute("userRole");
        if (currentUser != null) {
            if(currentUser.getType().equals("BUYER")) {
                //Loop BUYER Element of the .xml
                //if pageRequested.contains(value of the page at buyer element)             
                // chain.doFilter(request,response);
                // Else
                // Redirect the user to the main page
            }
            else if(currentUser.getType().equals("SELLER")) {
                //Same as above just for seller element
            }
            else if(currentUser.getType().equals("ADMINISTRATOR")) {
                //Same as above just for administrator element
            }           
        }
    }

    public void destroy() {
        // Not needed
    }
}

在doFilter方法内部的注释中解释了我需要做什么.有人可以给我一个提示,我应该如何遍历文件来查找每个用户类型的页面名称？我尝试从互联网上关注JAXP示例,但它们比我需要的更复杂.

更新
xml存储在WEB-INF / classes中

而是使用JAXB. JAXP是一个古老且非常详细的API. JAXB倾向于Javabeans,因此干净且相对容易.首先创建一个Javabean,它使用javax.xml.bind注释将1：1映射到XML文件.

@XmlRootElement
public class Access {

    @XmlElement
    private User buyer;

    @XmlElement
    private User seller;

    @XmlElement
    private User administrator;

    public User getBuyer() {
        return buyer;
    }

    public User getSeller() {
        return seller;
    }

    public User getAdministrator() {
        return administrator;
    }

    public static class User {

        @XmlElement(name="page")
        private List<String> pages;

        public List<String> getPages() {
            return pages;
        }

    }

}

然后执行以下部分来映射它(假设allowedpages.xml放在类路径的根目录中).

InputStream input = Thread.currentThread().getContextClassLoader().getResourceAsStream("allowedpages.xml");
Access access = (Access) JAXBContext.newInstance(Access.class).createUnmarshaller().unmarshal(input);

请注意,您不应该为此使用新的File().另见getResourceAsStream() vs FileInputStream.

最后,您可以访问所有买家页面,如下所示：

List<String> buyerPages = access.getBuyer().getPages();
// ...

毋庸置疑,养老保障并不总是最佳做法. Java EE 6附带容器管理的安全性.