jakarta-ee - 用户登录Web应用程序时如何实现 “Stay Logged In”

在大多数网站上，当用户要提供用户名和密码来登录系统时，会出现一个复选框，例如“保持登录状态”。如果选中此框，它将使您从同一Web浏览器跨所有 session 登录。如何在Java EE中实现相同功能？

我正在将基于FORM的容器管理的身份验证与JSF登录页面一起使用。

<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>

最佳答案

Java EE 8及更高版本
如果您使用的是Java EE 8或更高版本，请将 @RememberMe 与 HttpAuthenticationMechanism 一起放在自定义 RememberMeIdentityStore 上。

@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);
    }

}

您可以在a real world example中找到Java EE Kickoff Application。

Java EE 6/7
如果您使用的是Java EE 6或7，则在用户未登录但存在cookie的情况下，向其本地增长一个长期存在的cookie以跟踪唯一的客户端，并使用Servlet 3.0 API提供的编程性登录 HttpServletRequest#login() 。
如果您创建另一个数据库表，且其java.util.UUID值为PK，而所涉及的用户的ID为FK，则这是最容易实现的。
假定以下登录表单:

<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>

以下是映射到doPost()上的Servlet的/login方法中的以下内容:

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"，而COOKIE_AGE应该是以秒为单位的年龄，例如2592000为30天)
这是映射到受限页面上的doFilter()的Filter方法的样子:

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帮助程序方法结合使用(很遗憾，它们在Servlet API中丢失):

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);
}

尽管UUID非常难以暴力破解，但是您可以为用户提供一个选项，以将“记住”选项锁定到用户的IP地址(request.getRemoteAddr())，并将其存储/比较在数据库中。这使它更坚固。同样，将“到期日期”存储在数据库中将很有用。
无论用户何时更改密码，都应替换UUID值也是一个好习惯。

Java EE 5或更低版本
请升级。

关于jakarta-ee - 用户登录Web应用程序时如何实现 “Stay Logged In”，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/5082846/