如何实现“保持登录"当用户登录到 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,当用户要提供用户名和密码登录系统时,会出现保持登录状态"之类的复选框.如果您选中该框,它将让您在同一 Web 浏览器的所有会话中保持登录状态.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?
我将基于 FORM 的容器管理身份验证与 JSF 登录页面一起使用.
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 及更高版本
如果您使用的是 Java EE 8 或更新版本,请输入 @RememberMe
自定义 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 Java EE Kickoff 应用程序中的="noreferrer">真实示例一>.
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,自制一个长期存在的 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()
当用户未登录但存在 cookie 时.
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.
如果您创建另一个数据库表,其中 java.util.UUID
值为 PK,相关用户的 ID 为 FK,这是最容易实现的.
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>
以及映射到 /login
的 Servlet
的 doPost()
方法中的以下内容:
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"
并且 COOKIE_AGE
应该是秒,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)
以下是映射到受限页面的 Filter
的 doFilter()
方法的样子:
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);
}
结合那些 cookie helper 方法(可惜它们在 Servlet API 中没有):
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);
}
尽管 UUID
非常难以暴力破解,但您可以为用户提供一个选项来锁定记住"用户 IP 地址的选项 (request.getRemoteAddr()
) 并将其存储/比较到数据库中.这使它更加健壮.此外,具有到期日期"存储在数据库中会很有用.
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.
每当用户更改密码时替换 UUID
值也是一个好习惯.
It's also a good practice to replace the UUID
value whenever the user has changed its password.
请升级.
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