2015-12-26 35 views
1

在我的Java EE应用程序中,我通过JDBC领域实施了身份验证/自动化(第一次应用此解决方案)。使用JDBC领域进行身份验证

以下代码在成功登录时没有任何问题,问题是当我键入错误的凭据时:无论如何它都会登录,即使它捕获到ServletException(登录失败),这些代码行也不会执行在调试模式下):

request.setAttribute(“msg”,“Login in error”);

nextPage =“/errorPage.jsp”;

另一个奇怪的事情:无论什么我传递给

getServletContext()方法的getRequestDispatcher(下一页)的.forward(请求,响应 );

as nextPage(我试图把静态的“/errorPage.jsp”),它总是转发到index.jsp。

的Login.jsp

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

/** 
* @see HttpServlet#HttpServlet() 
*/ 
public Login() { 
    super(); 
    // TODO Auto-generated constructor stub 
} 

/** 
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response) 
*/ 
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { 
    // TODO Auto-generated method stub 
} 

/** 
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response) 
*/ 
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { 

    String username = request.getParameter("username").trim(); 
    String password = request.getParameter("password").trim(); 
    String nextPage = "/index.jsp"; 

    try { 
     request.login(username, password); 
    } 
    catch (ServletException ex) { 
     request.setAttribute("msg", "Error in login"); 
     nextPage = "/errorPage.jsp"; 
    } 

     getServletContext().getRequestDispatcher(nextPage).forward(request, response); 
    } 
} 

的login.jsp

<%@ page language="java" contentType="text/html; charset=UTF-8" 
    pageEncoding="UTF-8"%> 
<% 
    request.logout(); 
%> 
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> 
<html> 
<head> 
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> 
<title>Welcome</title> 
</head> 
<body> 
    <h1>Hi! You need to login.</h1> 
    <form method="POST" action="/MyApp/Login"> 
     Usuario: <input type="text" name="username" /> Password: <input 
      type="password" name="password" /> <input type="submit" 
      value="Send" /> 
    </form> 
</body> 
</html> 

的web.xml

<?xml version="1.0" encoding="UTF-8"?> 
<web-app xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" 
    version="3.0"> 
    <display-name>MyApp</display-name> 
    <welcome-file-list> 
     <welcome-file>login.jsp</welcome-file> 
    </welcome-file-list> 
    <login-config> 
     <auth-method>FORM</auth-method> 
     <realm-name>jdbcRealm</realm-name> 
     <form-login-config> 
      <form-login-page>/login.jsp</form-login-page> 
      <form-error-page>/errorPage.jsp</form-error-page> 
     </form-login-config> 
    </login-config> 
    <security-constraint> 
     <web-resource-collection> 
      <web-resource-name>Admin stuff</web-resource-name> 
      <url-pattern>/admin/*</url-pattern> 
      <http-method>GET</http-method> 
      <http-method>POST</http-method> 
     </web-resource-collection> 
     <auth-constraint> 
      <role-name>admin</role-name> 
     </auth-constraint> 
     <user-data-constraint> 
      <transport-guarantee>CONFIDENTIAL</transport-guarantee> 
     </user-data-constraint> 
    </security-constraint> 
    <security-constraint> 
     <web-resource-collection> 
      <web-resource-name>User stuff</web-resource-name> 
      <url-pattern>/user/*</url-pattern> 
      <http-method>GET</http-method> 
      <http-method>POST</http-method> 
     </web-resource-collection> 
     <auth-constraint> 
      <role-name>user</role-name> 
     </auth-constraint> 
     <user-data-constraint> 
      <transport-guarantee>CONFIDENTIAL</transport-guarantee> 
     </user-data-constraint> 
    </security-constraint> 
    <security-role> 
     <role-name>admin</role-name> 
    </security-role> 
    <security-role> 
     <role-name>user</role-name> 
    </security-role> 
</web-app> 

在此之前,我试过container-managed security溶液(与登录的表单动作调用j_security_check组件)。

登录正常工作与此(即使有错误的凭证),但我得到了另一个严重的问题之前,我没有:在一个用例,用户可以看到该项目正在开发,但它不应该能够看到其他用户的项目。我用下面的servlet实现了它,但问题在于(像其他解决方案一样),它跳过了一些指令(例如,在数据库中查找用户的指令),并且出现异常,重定向到错误页。

public class ViewUserProjects extends HttpServlet { 
    private static final long serialVersionUID = 1L; 

    public ViewUserProjects() { 
     super(); 
     // TODO Auto-generated constructor stub 
    } 

    protected void doGet(HttpServletRequest request, 
      HttpServletResponse response) throws ServletException, IOException { 
     DAO dao = (DAO) getServletContext().getAttribute("bd"); 

     Principal p = request.getUserPrincipal(); 
     String username = p.getName(); 

     try { 
      User user = dao.getUserByName(name); 
      request.getSession().setAttribute("user", user); 

      ArrayList<Project> projects = new ArrayList<Project>(); 

      tareas = ad.getUserProjects(Integer.parseInt(user.getId())); 

      request.setAttribute("projects", projects); 

      getServletContext().getRequestDispatcher(
        "/user/viewProjects.jsp").forward(request, 
          response); 
     } catch (Exception ex) { 
      request.setAttribute("msg", 
        "Error"); 
      getServletContext().getRequestDispatcher("/errorPage.jsp").forward(
        request, response); 
     } 
    } 

    protected void doPost(HttpServletRequest request, 
      HttpServletResponse response) throws ServletException, IOException { 
     // TODO Auto-generated method stub 
    } 

} 

回答

-1

你在哪里检查从表单输入中输入的用户名和密码是否正确。您是通过查询数据库进行身份验证还是通过匹配代码中设置的用户名和密码值进行身份验证?我相信这会给你一个线索。确保来自表单输入的用户名和密码与代码中设置的用户名和密码相匹配。让我知道你是否有其他问题。如果这个解决您的问题,请把它标记为正确答案

response.setContentType("text/html"); 
    String msg = " "; 

    String username = request.getParameter("username"); 
    String password = request.getParameter("password"); 
    try { 

if (username.equals("nick") && password.equals("nick_password")) { 


      nextPage = "HELLO" + username + "! Your login is SUCESSFULL"; 

     } else { 
      nextPage = "HELLO" + username + "!Your login is UNSUCESSFULL"; 
     } 

}// close try 
+0

认证在通过JDBC领域https://docs.oracle.com/javaee/6/tutorial/doc/glxgo.html – user3673449

1

当使用JDBCRealm,它是一个更好的做法是使用container-managed security应用authentication/authorization而不是从你的应用程序代码中处理这个(你正在做的)

所以我们允许服务器来处理这个问题,这意味着使用form-based authentication(您正在使用)按照Servlet Specification会是这样的:

一是形式:

<form action="j_security_check" method="POST"> 
    Username:<input type="text" name="j_username" placeholder="Username" /> 
    Password:<input type="password" name="j_password" placeholder="Password" /> 
    <input type="submit" value="Log In" /> 
</form> 

那么在我们Deployment Descriptor我们必须添加一些配置,你似乎已经有,但这里是另一个例子:

注意我相信你mising在这里我们使用了403错误* <error-page>标签这是Forbidden resource

<security-constraint> 
    <display-name>securityConstraint1</display-name> 
    <web-resource-collection> 
     <web-resource-name>resources</web-resource-name> 
     <description /> 
     <url-pattern>/protected/*</url-pattern> 
    </web-resource-collection> 
    <auth-constraint> 
     <role-name>appUser</role-name> 
     <role-name>appAdmin</role-name> 
    </auth-constraint> 
</security-constraint> 

<security-constraint> 
    <display-name>securityConstraint2</display-name> 
    <web-resource-collection> 
     <web-resource-name>resources</web-resource-name> 
     <description /> 
     <url-pattern>/protected/admni/*</url-pattern> 
    </web-resource-collection> 
    <auth-constraint> 
     <role-name>appAdmin</role-name> 
    </auth-constraint> 
</security-constraint> 

<login-config> 
    <auth-method>FORM</auth-method> 
    <realm-name>appRealm</realm-name> 
    <form-login-config> 
     <form-login-page>/index.xhtml</form-login-page> 
     <form-error-page>/public/forbidden.xhtml</form-error-page> 
    </form-login-config> 
</login-config> 

<security-role> 
    <role-name>appUser</role-name> 
</security-role> 

<security-role> 
    <role-name>appAdmin</role-name> 
</security-role> 

<error-page> 
    <error-code>403</error-code> 
    <location>/public/forbidden.xhtml</location> 
</error-page> 

我们不能忘了Data Protection(你已经拥有):

<user-data-constraint> 
    <transport-guarantee>CONFIDENTIAL</transport-guarantee> 
</user-data-constraint> 

所以现在我们需要定义ROLES的应用,这是在定义的应用程序服务器所以首先映射组来完成。 ¿您正在使用哪个应用程序服务器?

下面是使用GlassFish

我们需要添加一个例子的glassfish-web.xmlsun-web.xml

<?xml version="1.0" encoding="UTF-8"?> 
<!DOCTYPE sun-web-app PUBLIC "-//Sun Microsystems, Inc.//DTD GlassFish Application Server 3.0 Servlet 3.0//EN" "http://www.sun.com/software/appserver/dtds/sun-web-app_3_0-0.dtd"> 
<sun-web-app error-url=""> 

    <security-role-mapping> 
     <role-name>appUser</role-name> 
     <group-name>1</group-name> 
    </security-role-mapping> 

    <security-role-mapping> 
     <role-name>appAdmin</role-name> 
     <group-name>2</group-name> 
    </security-role-mapping> 

    <class-loader delegate="true"/> 
    <jsp-config> 
    <property name="keepgenerated" value="true"> 
     <description>Keep a copy of the generated servlet class' java code.</description> 
    </property> 
    </jsp-config> 
</sun-web-app> 

所以角色被映射到存在于real repository实际组名。 (这是直接在你的应用服务器上创建的)。

为了这个工作,我们需要在我们的DB中创建一个TABLE以定义用户组。

Realm此处创建于Server Admin Console

GlassFish中去:

配置>>服务器配置>>安全>>三界

而这里的境界配置的一个例子。

JDBCRealm

+0

谢谢您的回答来实现。在尝试我发布的解决方案之前,我先从你建议的那个开始。我用我遇到的问题更新了我的问题。 – user3673449

+0

您可以添加更多关于数据库如何映射的细节,您如何配置jdbcRealm以及您正在使用哪个应用程序服务器? –

+0

此外,编辑的代码似乎有点关闭,我没有看到你正在使用'tareas'或'username',因为你传递'dao.getUserByName(name)'我认为有些代码丢失了,你能请写完整的代码? –