下面的java spring引导代码表示对java后端的一个简单rest调用,该调用在mysql数据库上执行插入操作,但是当它执行rest调用时,我报告了下面的错误,我无法定义在mysql数据库中插入时的问题在何处,如何解决控制台中hibernate表示的错误?谢谢您
错误消息休眠:
"could not execute statement; SQL [n/a]; constraint [null];
nested exception is org.hibernate.exception.ConstraintViolationException:
could not execute statement
控制器:
@PostMapping("/newuser")
public User createUser(@Valid @RequestBody User userDetails) {
System.out.println("Sono nella creazione dell'utente");
return userRepository.save(userDetails);
}
型号:
@Entity
@Table(name = "users")
@EntityListeners(AuditingEntityListener.class)
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "id", unique = true, nullable = false)
private long id;
@Column(name = "first_name", nullable = false)
private String firstName;
@Column(name = "last_name", nullable = false)
private String lastName;
@Column(name = "email_address", nullable = true)
private String email;
/**
* Gets id.
*
* @return the id
*/
public long getId() {
return id;
}
/**
* Sets id.
*
* @param id the id
*/
public void setId(long id) {
this.id = id;
}
/**
* Gets first name.
*
* @return the first name
*/
public String getFirstName() {
return firstName;
}
/**
* Sets first name.
*
* @param firstName the first name
*/
public void setFirstName(String firstName) {
this.firstName = firstName;
}
/**
* Gets last name.
*
* @return the last name
*/
public String getLastName() {
return lastName;
}
/**
* Sets last name.
*
* @param lastName the last name
*/
public void setLastName(String lastName) {
this.lastName = lastName;
}
/**
* Gets email.
*
* @return the email
*/
public String getEmail() {
return email;
}
/**
* Sets email.
*
* @param email the email
*/
public void setEmail(String email) {
this.email = email;
}
/**
* Gets created at.
*
* @return the created at
*/
/**
* Sets created at.
*
* @param createdAt the created at
*/
/**
* Gets created by.
*
* @return the created by
*/
@Override
public String toString() {
return "User{" +
"id=" + id +
", firstName='" + firstName + '\'' +
", lastName='" + lastName + '\'' +
", email='" + email + '\'' +
'}';
}
}
存储库
import xxx.mode.User;
import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.stereotype.Repository;
/**
* The interface User repository.
*
* @author xxxx
*/
@Repository
public interface UserRepository extends JpaRepository<User, Long> {}
2条答案
按热度按时间kgsdhlau1#
出现休眠错误:
这只意味着您违反了数据库约束,并且在查看用户实体时,您有一些约束,例如firstname和lastname不应为null。因此,在执行下一行时,您将得到这些错误:
您确定这些字段不是空的吗?
你也在使用@valid注解,但我没有看到任何注解
javax.validation.constraints
因此,在调用api时,实际上不会进行验证。我可以向你推荐的是:使用
javax.validations
例如字段上的@notblank(message=“first name is required”)。这将在遇到api时抛出验证错误,并且不会继续userRepository.save(userDetails)
. 或者最好为您的请求体创建一个dto类,而不是使用实体-这是一种代码味道,但我猜这只是出于测试目的。此外,您还可以省略
private long id
字段,因为它已经用@id注解。主键已经是唯一的,不能修改。vulvrdjw2#
post对象中的guest firstname或lastname为空。因为您使用@valid注解验证对象,所以还要确保在反序列化期间在必需字段上添加@notnull来验证它们