Question

我有一个html表单，用于选择应存储在xml文件中的用户名和密码。这个存储用户名详细信息的类：

@XmlRootElement(name = "Emp")
@XmlType(propOrder = {"username", "passowrd"})

    public class UserBean {

        String username;
        String passowrd;

         @XmlElement(name = "username")
        public String getUsername() {
            return username;
        }

        public void setUsername(String username) {
            this.username = username;
        }

         @XmlElement(name = "password")
        public String getPassowrd() {
            return passowrd;
        }
        public void setPassowrd(String passowrd) {
            this.passowrd = passowrd;
        }


        @Override
        public String toString() {
            return "UserBean [username=" + username + ", passowrd=" + passowrd + "]";
        }

    }

这是我获取用户名详细信息的servlet，并尝试将它们存储在xml文件中：

@WebServlet("/Data")
public class Data extends HttpServlet {
    private static final long serialVersionUID = 1L;


    private static final String FILE_NAME = "jaxb-emp.xml";


    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

          UserBean emp = new  UserBean ();

         String userName = request.getParameter("username");
         String password = request.getParameter("password");

            emp.setUsername(userName);
            emp.setPassowrd(password);


             jaxbObjectToXML(emp);

             UserBean empFromFile = jaxbXMLToObject();
             System.out.println(empFromFile.toString());



         RequestDispatcher rd=request.getRequestDispatcher("index.jps");
         rd.forward(request, response);

    }



     private static UserBean jaxbXMLToObject() {
            try {
                JAXBContext context = JAXBContext.newInstance(UserBean.class);
                Unmarshaller un = context.createUnmarshaller();
                UserBean emp = (UserBean) un.unmarshal(new File(FILE_NAME));
                return emp;
            } catch (JAXBException e) {
                e.printStackTrace();
            }
            return null;
        }


        private static void jaxbObjectToXML(UserBean emp) {

            try {
                JAXBContext context = JAXBContext.newInstance(UserBean.class);
                Marshaller m = context.createMarshaller();
                //for pretty-print XML in JAXB
                m.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE);

                // Write to System.out for debugging
                // m.marshal(emp, System.out);

                // Write to File
                m.marshal(emp, new File(FILE_NAME));
            } catch (JAXBException e) {
                e.printStackTrace();
            }
        }
}

每当我启动tomcat并输入用户数据时，它都会覆盖该文件。问题是，我的目的是拥有许多用户名并避免重复。

Answer 1

您可以为商店UserBeans使用容器类。

算法：

将所有存储的bean读取到容器
如果没有用户名重复，则添加
将容器写入文件。

注意：将密码存储为plait文本是不安全的。你应该考虑安全问题。

容器：

@XmlRootElement(name = "Department")
@XmlAccessorType(value = XmlAccessType.FIELD)
public class DepartmentBean {

    @XmlElement(name = "Emp", type = UserBean.class)
    @XmlElementWrapper(name = "Emps")
    private List<UserBean> users;

    public List<UserBean> getUsers() {
        return users;
    }

    public void addIfAbsent(UserBean userBean) {
        if (users.stream().anyMatch(x -> x.getUsername().equals(userBean.getUsername()))) {
            System.err.println("User " + userBean + " is already exist, do nothing");
        } else {
            users.add(userBean);
        }
    }
}

xml示例（Department和Emps标记应始终位于文件中）：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<Department>
    <Emps>
        <Emp>
            <username>Name</username>
            <password>pass</password>
        </Emp>
    </Emps>
</Department>

使用示例：

public static void main(String... args) throws Exception {
    DepartmentBean departmentBean = read();
    UserBean user = new UserBean();
    user.setPassword("pass");
    user.setUsername("user2");

    departmentBean.addIfAbsent(user);
    write(departmentBean);
}

public static DepartmentBean read() throws JAXBException {
    JAXBContext context = JAXBContext.newInstance(DepartmentBean.class, UserBean.class);
    Unmarshaller unmarshaller = context.createUnmarshaller();
    return (DepartmentBean) unmarshaller.unmarshal(new StreamSource(new File("1.xml")));
}

public static void write(DepartmentBean department) throws JAXBException {
    JAXBContext context = JAXBContext.newInstance(DepartmentBean.class, UserBean.class);
    Marshaller marshaller = context.createMarshaller();
    marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE);
    marshaller.marshal(department, new File("1.xml"));
}

JAXB如何避免重复/覆盖对象

1 个答案: