如何实现“保持登录状态”当用户登录Web应用程序时 [英] How to implement "Stay Logged In" when user login in to the web application

问题描述

在大多数网站上，当用户即将提供登录系统的用户名和密码时，会出现一个保持登录状态的复选框。如果选中该复选框，它将使您在同一Web浏览器的所有会话中登录。如何在Java EE中实现相同的功能？

On most websites, when the user is about to provide the username and password to log into the system, there's a checkbox like "Stay logged in". If you check the box, it will keep you logged in across all sessions from the same web browser. How can I implement the same in Java EE?

我正在使用基于FORM的容器管理身份验证和JSF登录页面。

I'm using FORM based container managed authentication with a JSF login page.

<security-constraint>
    <display-name>Student</display-name>
    <web-resource-collection>
        <web-resource-name>CentralFeed</web-resource-name>
        <description/>
        <url-pattern>/CentralFeed.jsf</url-pattern>
    </web-resource-collection>        
    <auth-constraint>
        <description/>
        <role-name>STUDENT</role-name>
        <role-name>ADMINISTRATOR</role-name>
    </auth-constraint>
</security-constraint>
 <login-config>
    <auth-method>FORM</auth-method>
    <realm-name>jdbc-realm-scholar</realm-name>
    <form-login-config>
        <form-login-page>/index.jsf</form-login-page>
        <form-error-page>/LoginError.jsf</form-error-page>
    </form-login-config>
</login-config>
<security-role>
    <description>Admin who has ultimate power over everything</description>
    <role-name>ADMINISTRATOR</role-name>
</security-role>    
<security-role>
    <description>Participants of the social networking Bridgeye.com</description>
    <role-name>STUDENT</role-name>
</security-role>

推荐答案

Java EE 8及以上

如果您使用的是Java EE 8或更高版本，请输入 @RememberMe /javaee-spec/javadocs/javax/security/enterprise/authentication/mechanism/http/HttpAuthenticationMechanism.html\"rel =noreferrer> HttpAuthenticationMechanism 以及 RememberMeIdentityStore 。

@ApplicationScoped
@AutoApplySession
@RememberMe
public class CustomAuthenticationMechanism implements HttpAuthenticationMechanism {

    @Inject
    private IdentityStore identityStore;

    @Override
    public AuthenticationStatus validateRequest(HttpServletRequest request, HttpServletResponse response, HttpMessageContext context) {
        Credential credential = context.getAuthParameters().getCredential();

        if (credential != null) {
            return context.notifyContainerAboutLogin(identityStore.validate(credential));
        }
        else {
            return context.doNothing();
        }
    }
}

public class CustomIdentityStore implements RememberMeIdentityStore {

    @Inject
    private UserService userService; // This is your own EJB.

    @Inject
    private LoginTokenService loginTokenService; // This is your own EJB.

    @Override
    public CredentialValidationResult validate(RememberMeCredential credential) {
        Optional<User> user = userService.findByLoginToken(credential.getToken());
        if (user.isPresent()) {
            return new CredentialValidationResult(new CallerPrincipal(user.getEmail()));
        }
        else {
            return CredentialValidationResult.INVALID_RESULT;
        }
    }

    @Override
    public String generateLoginToken(CallerPrincipal callerPrincipal, Set<String> groups) {
        return loginTokenService.generateLoginToken(callerPrincipal.getName());
    }

    @Override
    public void removeLoginToken(String token) {
        loginTokenService.removeLoginToken(token);
    }

}

你可以找到一个真实世界的例子< Java EE Kickoff应用程序中的/ a>。

You can find a real world example in the Java EE Kickoff Application.

如果您使用的是Java EE 6或7， homegrow一个长期存在的cookie来跟踪唯一的客户端并使用Servlet 3.0 API提供的程序化登录 HttpServletRequest＃login（） -in但是cookie存在。

If you're on Java EE 6 or 7, homegrow a long-living cookie to track the unique client and use the Servlet 3.0 API provided programmatic login HttpServletRequest#login() when the user is not logged-in but the cookie is present.

如果你创建另一个带有 java.util.UUID的数据库表，这是最容易实现的。 code>作为PK的值和有问题的用户的ID作为FK。

This is the easiest to achieve if you create another DB table with a java.util.UUID value as PK and the ID of the user in question as FK.

假设以下登录表单：

<form action="login" method="post">
    <input type="text" name="username" />
    <input type="password" name="password" />
    <input type="checkbox" name="remember" value="true" />
    <input type="submit" />
</form>

以下 doPost（）方法一个 Servlet ，它映射在 / login ：

And the following in doPost() method of a Servlet which is mapped on /login:

String username = request.getParameter("username");
String password = hash(request.getParameter("password"));
boolean remember = "true".equals(request.getParameter("remember"));
User user = userService.find(username, password);

if (user != null) {
    request.login(user.getUsername(), user.getPassword()); // Password should already be the hashed variant.
    request.getSession().setAttribute("user", user);

    if (remember) {
        String uuid = UUID.randomUUID().toString();
        rememberMeService.save(uuid, user);
        addCookie(response, COOKIE_NAME, uuid, COOKIE_AGE);
    } else {
        rememberMeService.delete(user);
        removeCookie(response, COOKIE_NAME);
    }
}

（ COOKIE_NAME 应该是唯一的Cookie名称，例如remember和 COOKIE_AGE 应该是以秒为单位的年龄，例如 2592000 30天）

(the COOKIE_NAME should be the unique cookie name, e.g. "remember" and the COOKIE_AGE should be the age in seconds, e.g. 2592000 for 30 days)

以下是<$ c $的方式c> doFilter（）在受限制的页面上映射的过滤器的方法可能如下所示：

Here's how the doFilter() method of a Filter which is mapped on restricted pages could look like:

HttpServletRequest request = (HttpServletRequest) req;
HttpServletResponse response = (HttpServletResponse) res;
User user = request.getSession().getAttribute("user");

if (user == null) {
    String uuid = getCookieValue(request, COOKIE_NAME);

    if (uuid != null) {
        user = rememberMeService.find(uuid);

        if (user != null) {
            request.login(user.getUsername(), user.getPassword());
            request.getSession().setAttribute("user", user); // Login.
            addCookie(response, COOKIE_NAME, uuid, COOKIE_AGE); // Extends age.
        } else {
            removeCookie(response, COOKIE_NAME);
        }
    }
}

if (user == null) {
    response.sendRedirect("login");
} else {
    chain.doFilter(req, res);
}

与那些cookie帮助器方法相结合（太糟糕了，它们在Servlet API中缺失）：

In combination with those cookie helper methods (too bad they are missing in Servlet API):

public static String getCookieValue(HttpServletRequest request, String name) {
    Cookie[] cookies = request.getCookies();
    if (cookies != null) {
        for (Cookie cookie : cookies) {
            if (name.equals(cookie.getName())) {
                return cookie.getValue();
            }
        }
    }
    return null;
}

public static void addCookie(HttpServletResponse response, String name, String value, int maxAge) {
    Cookie cookie = new Cookie(name, value);
    cookie.setPath("/");
    cookie.setMaxAge(maxAge);
    response.addCookie(cookie);
}

public static void removeCookie(HttpServletResponse response, String name) {
    addCookie(response, name, null, 0);
}

虽然 UUID 非常难以蛮力，您可以为用户提供一个选项，将记住选项锁定到用户的IP地址（ request.getRemoteAddr（））并存储/比较它也在数据库中。这使得它更加强大。此外，在数据库中存储过期日期将非常有用。

Although the UUID is extremely hard to brute-force, you could provide the user an option to lock the "remember" option to user's IP address (request.getRemoteAddr()) and store/compare it in the database as well. This makes it a tad more robust. Also, having an "expiration date" stored in the database would be useful.

替换 UUID 值。

请升级。

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