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问题[spring-security]

Martin Hope
Дмитрий
Asked: 2023-12-10 00:45:12 +0000 UTC
  • 5

我正在使用 Spring Security 和 Thymeleaf 在 Spring Boot 中制作一个 Web 应用程序。视图是带有可以添加、查看、编辑和删除的评论的页面。我需要在视图中设置限制，以便在每个评论下的评论页面上，仅针对评论作者的用户显示“编辑评论”按钮，但我不知道如何在 Thymeleaf 中表达这一点。

我在网上找到了这些设计：

<div th:if="${#authorization.expression('hasRole(''ROLE_ADMIN'')')}">
  ADMIn section
</div>

<div sec:authentication="name">

<div sec:authentication="principal.authorities">.

但我不知道如何针对我的情况正确地重新表述这一点。谢谢。

spring-security
Martin Hope
V-CHO
Asked: 2022-07-26 14:55:04 +0000 UTC
  • 0

先生们，专家，告诉我，pliz：Spring Boot / Security 项目，当添加到 html/Thymeleaf sec:authentication=”principal.authorities”浏览器发出 500 和

Error retrieving value for property "”principal.authorities”" of authentication object of class org.springframework.security.authentication.UsernamePasswordAuthenticationToken

Invalid property '”principal' of bean class [org.springframework.security.authentication.UsernamePasswordAuthenticationToken]: Bean property '”principal' is not readable or has an invalid getter method: Does the return type of the getter match the parameter type of the setter?

同时，authentication.getAuthorities() 在控制台中显示值，一切都在调试中到位，如果您注释掉有问题的行，则访问限制适用于该按钮：

    <div sec:authorize="hasAuthority('ADMIN')">
    <form th:action="@{/bankdemo/accounts/show/{phone}(phone=${account.phone})}"
     th:method="delete">
        <input type="hidden" th:name="id" th:value="${bill.id}"/> 
            <button class="btn btn-danger"
            onclick="if (!(confirm('Are you sure you want to delete this bill?')))
                return false">
            Erase</button>
    </form>
    </div>

场地：

    @ElementCollection(targetClass = Role.class)
    @CollectionTable(name="roles", joinColumns = @JoinColumn(name="account_id"))
    @Enumerated(EnumType.STRING)
    private Set<Role> roles;

吸气剂：

    @Override
    public Collection<? extends GrantedAuthority> getAuthorities() {
        return roles;
    }

埃南：

public enum Role implements GrantedAuthority{
    ADMIN, CLIENT;

    @Override
    public String getAuthority() {
        return name();
    }
}

怎么了？

spring-boot spring-security
Martin Hope
Yan Zaitsau
Asked: 2022-07-13 23:37:56 +0000 UTC
  • 0

在这个视频的帮助下，我正在尝试制作一个类似的购物车。在这个阶段，我试图在用户的购物篮中获取有关产品（游览）的信息。已达到 35 分钟 在此处输入链接说明

我有一个连接旅游和用户的模型

@Entity 
@Table(name = "cart_items") public class CartItem {  
@Id 
@GeneratedValue(strategy = GenerationType.IDENTITY)  
private int id;
     
@ManyToOne  @JoinColumn(name = "tour_id")  
private Tour tour;
     
@ManyToOne  @JoinColumn(name = "user_id")  
private User user;
@Column(name = "order_date")  private Date date=new Date(); 
//getters and setters 
}

普通用户模型

package com.zaitsava.springboot_touristsite.entity;

import javax.persistence.*;
import java.util.Set;


@Entity
@Table(name = "user")
public class User {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;
    @Column(name = "firstname")
    private String firstname;

    @Column(name = "lastname")
    private String lastname;

    @Column(name = "patronymic")
    private String patronymic;

    @Column(name = "email")
    private String email;

    @Column(name = "phone")
    private String phone;

    @Column(name = "password")
    private String password;

    @Column(name = "active")
    private int active;

    @ManyToMany(cascade=CascadeType.ALL)
    @JoinTable(name="user_role", joinColumns=@JoinColumn(name="user_id"),
            inverseJoinColumns=@JoinColumn(name="role_id"))
    private Set<Role> roles;
    //gettes and setters
}

服务

import com.zaitsava.springboot_touristsite.entity.CartItem;
import com.zaitsava.springboot_touristsite.entity.User;
import com.zaitsava.springboot_touristsite.repository.CartItemRepository;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Service;

import java.util.List;

@Service
public class ShoppingCartService {
    @Autowired
    private CartItemRepository cartItemRepository;

    public List<CartItem> cartItemList(User user){
        return cartItemRepository.findByUser(user);
    };
}

和存储库：

package com.zaitsava.springboot_touristsite.repository;

import com.zaitsava.springboot_touristsite.entity.CartItem;
import com.zaitsava.springboot_touristsite.entity.User;
import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.stereotype.Repository;

import java.util.List;
@Repository
public interface CartItemRepository extends JpaRepository<CartItem,Integer> {
    public List<CartItem> findByUser(User user);
}

一切都通过测试很好地添加并写入数据库 在此处输入图像描述

说到产品显示控制器，问题就开始了。因为作者的实现用不同的方法扩展了User类，所以无法在控制器中获取当前用户。我试着做

package com.zaitsava.springboot_touristsite.controller;

import com.zaitsava.springboot_touristsite.entity.CartItem;
import com.zaitsava.springboot_touristsite.entity.CurrentUser;
import com.zaitsava.springboot_touristsite.entity.User;
import com.zaitsava.springboot_touristsite.service.ShoppingCartService;
import com.zaitsava.springboot_touristsite.service.UserServiceImpl;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.web.bind.annotation.GetMapping;


import java.util.List;

@Controller
public class ShoppingCartController {
    @Autowired
    private ShoppingCartService cartService;
    @Autowired
    private UserServiceImpl userService;

    @GetMapping("/cart")
    public String showCart(Model model,@CurrentUser User user){

        if(user==null) System.out.println("User is null");

        List<CartItem> cartItemList=cartService.cartItemList(user);
        model.addAttribute("cartItems",cartItemList);
        return "user/cart";
    }


}

通过当前用户的注解

package com.zaitsava.springboot_touristsite.entity;
import org.springframework.security.core.annotation.AuthenticationPrincipal;
import java.lang.annotation.*;

@Target({ElementType.PARAMETER, ElementType.TYPE})
@Retention(RetentionPolicy.RUNTIME)
@Documented
@AuthenticationPrincipal
public @interface CurrentUser {}

但是，我得到一个空用户，即使我通过 id 为 4 的用户登录。我也尝试通过参数

 public String showCart(Model model,Principal principal){
        Authentication authentication = (Authentication) principal;
        User user= (User) authentication.getPrincipal();
        //остальной код
}

但后来我收到有关不正确类型转换的错误

更新 了添加的 Spring Security 设置

@Configuration
@EnableWebSecurity
public class SecurityConfiguration extends WebSecurityConfigurerAdapter {

    @Autowired
    private BCryptPasswordEncoder bCryptPasswordEncoder;

    @Autowired
    private DataSource dataSource;

    private final String USERS_QUERY = "select email, password, active from user where email=?";
    private final String ROLES_QUERY = "select u.email, r.role from user u inner join user_role ur on (u.id = ur.user_id) inner join role r on (ur.role_id=r.role_id) where u.email=?";

    @Override
    protected void configure(AuthenticationManagerBuilder auth) throws Exception {
        auth.jdbcAuthentication()
                .usersByUsernameQuery(USERS_QUERY)
                .authoritiesByUsernameQuery(ROLES_QUERY)
                .dataSource(dataSource)
                .passwordEncoder(bCryptPasswordEncoder);
    }

    @Override
    protected void configure(HttpSecurity http) throws Exception {
        http.authorizeRequests()
                .antMatchers("/").permitAll()
                .antMatchers("/user/cart").authenticated()
                .antMatchers("/login").permitAll()
                .antMatchers("/signup").permitAll()
                .antMatchers("favicon.ico").permitAll()
               /* .antMatchers("/admin/**").access("hasRole('ADMIN')")*/
                .antMatchers("/main/**").hasAuthority("ADMIN").anyRequest()
                .authenticated().and().csrf().disable()
                .formLogin().loginPage("/login").permitAll().failureUrl("/login?error=true")
                .defaultSuccessUrl("/")
                .usernameParameter("email")
                .passwordParameter("password")
                .and().logout()
                .logoutRequestMatcher(new AntPathRequestMatcher("/logout"))
                .logoutSuccessUrl("/")
                .and().rememberMe()
                .tokenRepository(persistentTokenRepository())
                .tokenValiditySeconds(60 * 60)
                .and().exceptionHandling().accessDeniedPage("/access_denied");

    }

    @Override
    public void configure(WebSecurity web) throws Exception {
        web.ignoring()
                .antMatchers("/images/**")
                .antMatchers("/fonts/**");
    }

    @Bean
    public PersistentTokenRepository persistentTokenRepository() {
        JdbcTokenRepositoryImpl db = new JdbcTokenRepositoryImpl();
        db.setDataSource(dataSource);

        return db;
    }
}

用户控制器：

package com.zaitsava.springboot_touristsite.controller;


import javax.validation.Valid;

import com.zaitsava.springboot_touristsite.entity.User;
import com.zaitsava.springboot_touristsite.service.UserService;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.security.core.Authentication;
import org.springframework.security.core.context.SecurityContextHolder;
import org.springframework.stereotype.Controller;
import org.springframework.validation.BindingResult;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.PostMapping;
import org.springframework.web.servlet.ModelAndView;

@Controller
public class UserController {

    @Autowired
    private UserService userService;

    @GetMapping("/signup")
    public ModelAndView signup() {
        ModelAndView model = new ModelAndView();
        User user = new User();
        model.addObject("user", user);
        model.setViewName("user/signup");
        return model;
    }

    @PostMapping("/signup")
    public ModelAndView createUser(@Valid User user, BindingResult bindingResult) {
        ModelAndView model = new ModelAndView();
        User userExists = userService.findUserByEmail(user.getEmail());

        if(userExists != null) {
            bindingResult.rejectValue("email", "error.user", "User with this email exists");
        }
        if(bindingResult.hasErrors()) {
            model.setViewName("user/signup");
        } else {
            userService.saveUser(user);
            model.addObject("msg", "User succesful register!");
            model.addObject("user", new User());
            model.setViewName("redirect:home/main");
        }

        return model;
    }



    @GetMapping("/home/main")
    public ModelAndView home() {
        ModelAndView model = new ModelAndView();
        Authentication auth = SecurityContextHolder.getContext().getAuthentication();
        User user = userService.findUserByEmail(auth.getName());

        model.addObject("userName", "Hello,"+user.getFirstname() + " " + user.getLastname()+" "+user.getPatronymic());
        model.setViewName("redirect:/");
        return model;
    }

    @GetMapping("/access_denied")
    public ModelAndView accessDenied() {
        ModelAndView model = new ModelAndView();
        model.setViewName("errors/access_denied");
        return model;
    }
}

我不明白我需要做什么才能让控制器正常工作

spring-boot spring-security
Martin Hope
AlexPopov
Asked: 2021-10-24 00:55:59 +0000 UTC
  • 0

目前，通过html表单向数据库添加新用户时，密码以我在表单中指定的方式写入数据库，但是是否有可能以某种方式使密码立即以加密形式进入数据库？

在用 spring MVC + security + hibernate + thymeleaf 编写的简单 crud 应用程序中组织这个的最佳方法是什么？

大部分例子都是用xml文件写的（或者我没好好搜索），但目前我不知道如何在javaConfig上搞砸这个东西。

链接到git 存储库

BCryptPasswordEncoder bean 和自动装配使用它

    @Bean
    public BCryptPasswordEncoder bCryptPasswordEncoder() {
        return new BCryptPasswordEncoder(12);
    }

    @Autowired
    protected void configureGlobal(AuthenticationManagerBuilder auth) throws Exception {
        auth.userDetailsService(userService).passwordEncoder(bCryptPasswordEncoder());
    }

添加用户的html表单

<form action="#" th:action="@{addUser}" th:object="${user}" method="post">
    <label for="name">Name</label>
    <input type="text" th:field="*{name}" id="name" placeholder="Name">
    <span th:if="${#fields.hasErrors('name')}" th:errors="*{name}"></span>

    <label for="password">Password</label>
    <input type="text" th:field="*{password}" id="password" placeholder="Password">
    <span th:if="${#fields.hasErrors('password')}" th:errors="*{password}"></span>

    <label for="email">Email</label>
    <input type="text" th:field="*{email}" id="email" placeholder="Email">
    <span th:if="${#fields.hasErrors('email')}" th:errors="*{email}"></span>
    
    <input type="submit" value="Add User">
</form>

添加新用户的控制器

    @GetMapping
    public ModelAndView showRegistrationForm(User user) {
        ModelAndView modelAndView = new ModelAndView("reg");
        modelAndView.addObject("user", user);
        return modelAndView;
    }

    @PostMapping("/newUser")
    public String createUser(User user) {
        user.getRoleSet().add(roleService.getDefaultRole());
        userService.addUser(user);
        return "redirect:/index";
    }

ps addUser 方法调用服务上的 addUser 方法，该方法又使用 JPA (save(user)) 实现

spring-security
Martin Hope
bsuart
Asked: 2020-09-19 05:36:20 +0000 UTC
  • 3

我是否正确理解了 jwt 在微服务架构中的使用：

在身份验证期间，会为用户创建访问令牌和刷新令牌。刷新令牌存储访问令牌的 id。访问令牌在标头中发送给用户并存储在 WebStorage 或 Cookies 中，而访问令牌存储在某处（数据库或文件）。在服务器上客户端用户的后续请求中，访问令牌被发送到服务器，在那里检查其有效性，然后，如果成功，则检查生命周期，如果有，则发出新的访问令牌。

spring-security