Java Spring Security中需要授权的文本表示

我在Spring Security中的注册权限有问题 我不能做注册的方法

我试图设置每个路径的访问权限，但没有帮助

控制器

证券配置

用户服务

用户模型

错误

java.lang.IllegalArgumentException：授予的文本权限 代表须于 org.springframework.util.Assert.hasTextAssert.java:284 ~[spring-core-5.2.4.释放。震击器：5.2.4.释放]位于 org.springframework.security.core.authority.SimpleGrantedAuthority.SimpleGrantedAuthority.java:38 ~[spring-security-core-5.2.2.RELEASE.jar:5.2.2.RELEASE]at com.xxx.xx.models.User.getAuthoritiesUser.java:71~[classes/:na] java.base/jdk.internal.reflect.NativeMethodAccessorImpl.invoke0Native 方法~[na:na]at java.base/jdk.internal.reflect.NativeMethodAccessorImpl.invokeNativeMethodAccessorImpl.java:62 ~[na:na]at java.base/jdk.internal.reflect.DelegatingMethodAccessorImpl.invokeDelegatingMethodAccessorImpl.java:43 ~[na:na]at java.base/java.lang.reflect.Method.invokeMethod.java:566~[na:na]

---

在用户模型类中，确保设置了角色，以便getAuthories方法能够工作

错误提示您正在使用空角色执行新的SimpleGrantedAuthority

 @Override
    public Collection<? extends GrantedAuthority> getAuthorities() {
        List<GrantedAuthority> listRole = new ArrayList<GrantedAuthority>();

        listRole.add(new SimpleGrantedAuthority(role)); // this is the problematic line!
        return listRole;
    }

如果您没有角色，那么只需返回一个空列表即可

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

---

这是我的JSON，我使用defult角色。 我试图返回一个空列表，但我有同样的错误

{
  "accountNonExpired": true,
  "accountNonLocked": true,
  "authorities": [
    {
      "authority": "string"
    }
  ],
  "credentialsNonExpired": true,
  "enabled": true,
  "id": 0,
  "lastName": "Piotr",
  "name": "Piotr",
  "password": "Piotr",
  "role": "ROLE_ADMIN",
  "username": "Piotr"
}

邮差表格错误

**"timestamp": "2020-04-01T15:20:05.670+0000",
    "status": 500,
    "error": "Internal Server Error",
    "message": "JSON conversion problem: A granted authority textual representation is required; nested exception is com.fasterxml.jackson.databind.JsonMappingException: A granted authority textual representation is required\n at [Source: (PushbackInputStream); line: 4, column: 18] (through reference chain: com.xxx`enter code here`.xxx.models.User[\"authorities\"])",
    "path": "/register"
}**

---

我相信你必须创建用户对象，比如

新的org.springframework.security.core.userdetails.Usersername，passwordencode.encode user.getPass，grantedAuthorityList

这是我的作品

@Service  
public class UserSecurityService implements UserDetailsService {

@Autowired
YourRepository yourRepository;


@Autowired
PasswordEncoder passwordEncoder;


@Override
public UserDetails loadUserByUsername(String username) throws UsernameNotFoundException {

    List<GrantedAuthority> grantedAuthorityList = new ArrayList<>();
    User user = yourRepository.findUserByUser(username);

    List<String> listRoles = new ArrayList<>();
    //List<UserRoles> userRoles = user.getUserRoleList();

    user.getUserRoleList().forEach(role->listRoles.add(role.getRole().getRole()));// get role from database - usa il tuo modo **

    grantedAuthorityList = listRoles.stream().map(SimpleGrantedAuthority::new).collect(Collectors.toList());


    //*** important ****
    return new org.springframework.security.core.userdetails.User(username,
            passwordEncoder.encode(user.getPass()), grantedAuthorityList);


}

---

请参阅我的最新答案。对于空列表，它肯定会起作用，因为您不再使用新的SimpleGrantedAuthority，这就是您的错误所在。也许错误与此不同？我发现错误是JSON的错误，因为我使用了swagger，他给我生成了一个错误的JsonI：在[Source:PushbackInputStream；line:5，column:5]通过引用链：models.User[Authories]->java.util.Collections$EmptyList[0]]对于根原因com.fasterxml.jackson.databind.exc.InvalidDefinitionException:无法构造org.springframework.security.core.GrantedAuthority的实例与默认构造一样，不存在任何创建者：抽象类型需要映射到具体类型，具有自定义反序列化器，或者在[来源：PushbackInputStream；行：5，列：5]通过引用链：models.User[authorities]->java.util.Collections$EmptyList[0]

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

{
  "accountNonExpired": true,
  "accountNonLocked": true,
  "authorities": [
    {
      "authority": "string"
    }
  ],
  "credentialsNonExpired": true,
  "enabled": true,
  "id": 0,
  "lastName": "Piotr",
  "name": "Piotr",
  "password": "Piotr",
  "role": "ROLE_ADMIN",
  "username": "Piotr"
}

**"timestamp": "2020-04-01T15:20:05.670+0000",
    "status": 500,
    "error": "Internal Server Error",
    "message": "JSON conversion problem: A granted authority textual representation is required; nested exception is com.fasterxml.jackson.databind.JsonMappingException: A granted authority textual representation is required\n at [Source: (PushbackInputStream); line: 4, column: 18] (through reference chain: com.xxx`enter code here`.xxx.models.User[\"authorities\"])",
    "path": "/register"
}**

@Service  
public class UserSecurityService implements UserDetailsService {

@Autowired
YourRepository yourRepository;


@Autowired
PasswordEncoder passwordEncoder;


@Override
public UserDetails loadUserByUsername(String username) throws UsernameNotFoundException {

    List<GrantedAuthority> grantedAuthorityList = new ArrayList<>();
    User user = yourRepository.findUserByUser(username);

    List<String> listRoles = new ArrayList<>();
    //List<UserRoles> userRoles = user.getUserRoleList();

    user.getUserRoleList().forEach(role->listRoles.add(role.getRole().getRole()));// get role from database - usa il tuo modo **

    grantedAuthorityList = listRoles.stream().map(SimpleGrantedAuthority::new).collect(Collectors.toList());


    //*** important ****
    return new org.springframework.security.core.userdetails.User(username,
            passwordEncoder.encode(user.getPass()), grantedAuthorityList);


}