Question

我是一个尝试学习JSF的新手。你可以帮我解决日食中的错误404！我尝试通过进入属性更改服务器详细信息，但它仍然给我错误。

请帮我解决这个问题。提前谢谢。

我的登录代码如下：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" 
      xmlns:ui="http://java.sun.com/jsf/facelets" 
      xmlns:h="http://java.sun.com/jsf/html" 
      xmlns:f="http://java.sun.com/jsf/core">

   <head><title>JSF Login</title></head>
 <body>
     <h1>Login</h1>
 <h:form>
<table>
 <tr>
<td><h:outputText value="Username: " /></td>
<td><h:inputText id="loginname" 
 value="#{loginBean.uname}" />
 </td>
</tr>
<tr>
<td><h:outputText value="Password: " /></td>
<td><h:inputSecret id="password" 
value="#{loginBean.password}" />
</td>
</tr>
<tr>
<td> </td>
<td><h:commandButton value="Login" 
action="#{loginBean.loginProject()}"/>
</td>
</tr>

</h:form>
</body>
</html>

我的loginbean文件：

package beans;

import dao.UserDAO;
import java.io.Serializable;
import javax.faces.application.FacesMessage;
import javax.faces.bean.ManagedBean;
import javax.faces.bean.SessionScoped;
import javax.faces.context.FacesContext;
import javax.servlet.http.HttpSession;

@ManagedBean(name = "loginBean")
@SessionScoped
/**
 *
 * @author User
 */
public class LoginBean implements Serializable {

    private static final long serialVersionUID = 1L;
    private String password;
    private String message, uname;

    public String getMessage() {
        return message;
    }

    public void setMessage(String message) {
        this.message = message;
    }

    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    public String getUname() {
        return uname;
    }

    public void setUname(String uname) {
        this.uname = uname;
    }

    public String loginProject() {
        boolean result = UserDAO.login(uname, password);
        if (result) {
            // get Http Session and store username
            HttpSession session = Util.getSession();
            session.setAttribute("username", uname);

            return "home";
        } else {

            FacesContext.getCurrentInstance().addMessage(
                    null,
                    new FacesMessage(FacesMessage.SEVERITY_WARN,
                    "Invalid Login!",
                    "Please Try Again!"));

            // invalidate session, and redirect to other pages

            //message = "Invalid Login. Please Try Again!";
            return "login";
        }
    }

    public String logout() {
      HttpSession session = Util.getSession();
      session.invalidate();
      return "login";
   }
}

Web.xml文件：

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd" id="WebApp_ID" version="3.1">
 <context-param>
  <param-name>javax.faces.PROJECT_STAGE</param-name>
  <param-value>Development</param-value>
 </context-param>
  <listener>  
        <listener-class>  
            com.sun.faces.config.ConfigureListener  
        </listener-class>  
    </listener>
 <welcome-file-list>
  <welcome-file>login.xhtml</welcome-file>
 </welcome-file-list>

</web-app>

Answer 1

假设存在所有依赖项和所有其他配置，您只需将以下内容添加到web.xml即可。我还建议您阅读java ee 7教程并使用Netbeans，因为您刚刚开始。

 <servlet>
    <servlet-name>Faces Servlet</servlet-name>
    <servlet-class>javax.faces.webapp.FacesServlet</servlet-class>
    <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>Faces Servlet</servlet-name>
    <url-pattern>*.xthml</url-pattern>
  </servlet-mapping>

Eclipse Tomcat错误404

1 个答案: