我是这个主题的新手,在尝试进行授权时,我会不断收到“ java.lang.StackOverflowError:null”。我在邮递员中收到内部服务器错误,在控制台中我得到了stackoveflowerror。这是我的AuthenticationController:
@PostMapping("/signup")
public ResponseEntity<?> registerUser(@Valid @RequestBody SignUpRequest signUpRequest) {
if (userService.existsByEmail(signUpRequest.getEmail())) {
throw new BadRequestException("email already exists");
}
User user = UserMapper.INSTANCE.registerRequestoUser(signUpRequest);
user.setPassword(passwordEncoder.encode(user.getPassword()));
Optional<Role> optionalRole = roleService.getByName("user");
if (optionalRole.isPresent()) {
Role userRole = optionalRole.get();
user.addRole(userRole);
Optional<User> optionalUser = userService.create(user);
if (optionalUser.isPresent()) {
User result = optionalUser.get();
URI location = ServletUriComponentsBuilder
.fromCurrentContextPath().path("/api/v1/users/email/{email}")
.buildAndExpand(result.getEmail()).toUri();
return ResponseEntity.created(location).body("User registered successfully");
}
}
return (ResponseEntity<?>) ResponseEntity.badRequest();
}
我的SignUpRequest类:
public class SignUpRequest {
@NotBlank
private String firstName;
@NotBlank
private String lastName;
@NotBlank
@Email
private String email;
@NotBlank
@Size(min = 8, max = 20)
private String password;
private Set<String> roles;
我的用户实体和我的角色实体:
@Entity
@Data
@NoArgsConstructor
@AllArgsConstructor
public class User extends BaseEntity {
private String firstName;
private String lastName;
@Email(message = "Email should be valid")
private String email;
@Size(min = 3, max = 100, message
= "password must be between 3 and 50 characters")
private String password;
@OneToMany(mappedBy = "user",
cascade = {CascadeType.PERSIST, CascadeType.MERGE,
CascadeType.DETACH, CascadeType.REFRESH})
private Set<Appointment> appointments;
@ManyToMany(targetEntity = Role.class,
cascade = {CascadeType.ALL},
fetch = FetchType.EAGER)
@JoinTable(
name = "user_role",
joinColumns = {@JoinColumn(name = "user_id")},
inverseJoinColumns = {@JoinColumn(name = "role_id")}
)
private Set<Role> roles;
@Entity
@Data
@NoArgsConstructor
@AllArgsConstructor
public class Role extends BaseEntity {
@NotNull
private String name;
@ManyToMany(mappedBy = "roles")
private Set<User> users;
public String getName() {
return name;
}
这是我的邮递员要求
{
"firstName":"name",
"lastName":"name",
"email": "[email protected]",
"password": "thepassword123",
"roles": ["user"]
}
这里是堆栈跟踪的一部分:
at at com.project.rushhour.entity.User.hashCode(User.java:15) ~[classes/:na]
...
at at com.project.rushhour.entity.Role.hashCode(Role.java:15) ~[classes/:na]
...
at com.project.rushhour.entity.User.hashCode(User.java:15) ~[classes/:na]
...
at com.project.rushhour.entity.Role.hashCode(Role.java:15) ~[classes/:na]
...
at com.project.rushhour.entity.User.hashCode(User.java:15) ~[classes/:na]
...
at com.project.rushhour.entity.Role.hashCode(Role.java:15) ~[classes/:na]
...
我认为程序尝试解析角色时,该错误在某处,但不确定到底是什么问题。 jwt的东西设置正确,只是逻辑错了
问题的原因是,您具有从用户到角色和从角色到用户的关系。但是您尚未显式定义hashCode方法。您使用@ Data批注。默认情况下,Lombok生成的hashCode方法包括all实体的属性。
hashCode通常如下计算:
h = hashCode(property1);
h = h*31 + hashCode(property2);
h = h*31 + hashCode(property3);
h = h*31 + hashCode(property4);
h = h*31 + hashCode(property5);
...
在User上调用hashCode时,如果计算属性firstName,lastName等的哈希码,则还要添加属性roles]的哈希码>。 roles的哈希码的计算包括每个Role元素的哈希码的计算。
但是哈希码Role
的计算包括属性users的哈希码的计算,该属性在每个User元素上调用hashCode()。正如我们在上面看到的,User的哈希码将再次与Role的哈希码冲突。等。直到使用堆栈容量。然后您得到StackOverflowError。如何避免?从hashCode()
方法中排除递归。要么手动实现hashCode()。或告诉Lombok从hashCode()中排除基于关系的属性。我希望龙目岛仍然可以为我产生一切。因此,我会告诉Lombok在生成hashCode()和equals( )
。因此您的代码如下所示:public class User extends BaseEntity {
private String firstName;
private String lastName;
...
@EqualsAndHashCode.Exclude
private Set<Role> roles;
...
}
public class Role extends BaseEntity {
...
private String name;
...
@EqualsAndHashCode.Exclude
private Set<User> users;
...
}
为什么注释称为EqualsAndHashCode?因为hashCode()和equals()应该始终进行[[synchronously更改,以实现所谓的
hashCode-equals
合同。