Java 如何在所有模板中显示当前登录用户的信息，包括在 Spring Security 应用程序中由 WebMvcConfigurerAdapter 管理的视图
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How to display current logged-in user's information in all templates including view managed by WebMvcConfigurerAdapter in Spring Security application
提问by Dee
I have a Spring Boot application that uses Spring Security and Thymeleaf template. I am trying to display the logged-in user's first name and last name in a template when the controller is managed by a subclass of WebConfigurerAdapter.
我有一个使用 Spring Security 和 Thymeleaf 模板的 Spring Boot 应用程序。当控制器由 WebConfigurerAdapter 的子类管理时，我试图在模板中显示登录用户的名字和姓氏。
So, say my WebConfigurerAdapter subclass looks like this
所以，说我的 WebConfigurerAdapter 子类看起来像这样
@Configuration
public class MvcConfig extends WebMvcConfigurerAdapter{
@Override
public void addViewControllers(ViewControllerRegistry registry){
registry.addViewController("/some-logged-in-page").setViewName("some-logged-in-page");
registry.addViewController("/login").setViewName("login");
}
....
}
My User entity class looks like this
我的 User 实体类看起来像这样
@Entity
@Table(name = "user")
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "id", nullable = false, updatable = false)
private Long id;
@Column(name="first_name", nullable = false)
private String firstName;
public String getFirstName() {
return firstName;
}
...
}
In my template, I have tried using code like
在我的模板中，我尝试使用类似的代码
<div sec:authentication="firstName"></div>
But it didn't work.
但它没有用。
I know it is possible to use a ControllerAdvise as follows:
我知道可以使用 ControllerAdvise 如下：
@ControllerAdvice
public class CurrentUserControllerAdvice {
@ModelAttribute("currentUser")
public UserDetails getCurrentUser(Authentication authentication) {
return (authentication == null) ? null : (UserDetails) authentication.getPrincipal();
}
}
and then access the details in the template using code like:
然后使用以下代码访问模板中的详细信息：
<span th:text ="${currentUser.getUser().getFirstName()}"></span>
But this doesn't work with any view controller registered with my class MvcConfig. Rather I will need to make sure each of my controllers are separate classes.
但这不适用于向我的类 MvcConfig 注册的任何视图控制器。相反，我需要确保我的每个控制器都是单独的类。
So, could someone kindly point me to a way to automatically insert the logged-in user details to my view, e.g. some-logged-in-page.html in this example? Thanks
那么，有人可以向我指出一种自动将登录用户详细信息插入我的视图的方法，例如本示例中的 some-logged-in-page.html 吗？谢谢
采纳答案by Dee
It's quite easy to accomplish this, thanks to a hint from Balaji Krishnan.
多亏了 Balaji Krishnan 的提示，这很容易实现。
Basically, I had to add the Thymeleaf Spring Security integration module to my build.gradle file as follows:
基本上，我必须将 Thymeleaf Spring Security 集成模块添加到我的 build.gradle 文件中，如下所示：
compile("org.thymeleaf.extras:thymeleaf-extras-springsecurity3")
Then in my template I just used the following markup:
然后在我的模板中，我只使用了以下标记：
<span th:text ="${#authentication.getPrincipal().getUser().getFirstName()}"></span>
回答by B378
When using Spring Security 4 and Thymeleaf 3:
使用 Spring Security 4 和 Thymeleaf 3 时：
<span th:text="${#authentication.getPrincipal().getUsername()}"></span>
回答by Thomas Lang
This construct is working for me (spring boot 1.5 and 2.0/thymeleaf 3):
It is documented here (bottom of the page) Thymeleaf + Spring Security integration basics
此构造对我有用（spring boot 1.5 和 2.0/thymeleaf 3）：
此处记录（页面底部）Thymeleaf + Spring Security 集成基础知识
Logged user: <span sec:authentication="name">Bob</span>
Don′t forget to include the sec tag in the html section of your view:
不要忘记在视图的 html 部分包含 sec 标记：
<!DOCTYPE html>
<html xmlns:th="http://www.thymeleaf.org"
xmlns:sec="http://www.thymeleaf.org/extras/spring-security">
<head>
</head>
<body>
I hope this helps!
Have a nice day!
Thomas
我希望这有帮助！
祝你今天过得愉快！
托马斯
回答by Tanaji Kolekar
As @B378 said, When using Spring Security 4 and Thymeleaf 3: you have to use following.
正如@B378 所说，使用 Spring Security 4 和 Thymeleaf 3 时：您必须使用以下内容。
<span th:text="${#authentication.getPrincipal().getUsername()}"></span>
Because spring security uses UserDetails internally. And UserDetails contains one function called getUsername().
因为 spring security 在内部使用 UserDetails。UserDetails 包含一个名为 getUsername() 的函数。
回答by sunitkatkar
回答by user1145691
For me, When using Spring boot 2.1.2 I need to use the following
对我来说，使用 Spring boot 2.1.2 时，我需要使用以下内容
<span th:text="${#authentication.getPrincipal()}"></span> <!-- No ".getUsername()"-->
With thymeleaf-extras-springsecurity5
使用 thymeleaf-extras-springsecurity5
回答by user1677230
When using Spring boot 2.2.1.
使用 Spring Boot 2.2.1 时。
For the maven, Add these lines to the pom.xml
对于 maven，将这些行添加到 pom.xml
<dependency>
<groupId>org.thymeleaf.extras</groupId>
<artifactId>thymeleaf-extras-springsecurity5</artifactId>
</dependency>
<dependency>
<groupId>org.thymeleaf.extras</groupId>
<artifactId>thymeleaf-extras-springsecurity5</artifactId>
</dependency>
In the thymeleaf
在百里香叶
<span th:text="${#authentication.getPrincipal().getUsername()}"></span>
<span th:text="${#authentication.getPrincipal().authorities}"></span>
<span th:text="${#authentication.getPrincipal().getUsername()}"></span>
<span th:text="${#authentication.getPrincipal().authorities}"></span>
回答by Ullas Hunka
For thymleaf 4 and above you have to modify some classes in order to get the info. Heres how you can do that:
对于 thymleaf 4 及更高版本，您必须修改一些类才能获取信息。以下是您如何做到这一点：
First, add the getter of your user getUser(){return this.user;}in the UserDetails class and then push that object in the UserDetailsService object. Without these changes, thymleaf will not parse your HTML file.
首先，getUser(){return this.user;}在 UserDetails 类中添加用户的 getter，然后将该对象推送到 UserDetailsService 对象中。如果没有这些更改，thymleaf 将不会解析您的 HTML 文件。
Then you can get the info as follows:
然后你可以得到如下信息：
<span sec:authentication="principal.user.name">
<span sec:authentication="principal.user.name">
