【hibernate】主键生成策略使用UUID报出如下警告：org.hibernate.id.UUIDHexGenerator - HHH000409: Using org.hibernate.id.UUIDHexGenerator which does not generate IETF RFC

主键生成策略使用UUID报出如警告如下：

控制台- 2017-11-24 18:40:14 [restartedMain] WARN  org.hibernate.id.UUIDHexGenerator - HHH000409: Using org.hibernate.id.UUIDHexGenerator which does not generate IETF RFC 4122 compliant UUID values; consider using org.hibernate.id.UUIDGenerator instead

说是它不生成符合IETF RFC 4122标准的UUID值; 请考虑使用org.hibernate.id.UUIDGenerator。

实体如下：

package com.sxd.entity;

import org.hibernate.annotations.GenericGenerator;

import javax.persistence.*;

@Entity
@GenericGenerator(name = "jpa-uuid", strategy = "uuid")
public class User {
    private String id;
    private String username;
    private String password;
    private Integer age;

    @Id
    @GeneratedValue(generator = "jpa-uuid")
    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    @Column(nullable = false)
    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    @Column(nullable = false)
    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    @Column(nullable = false)
    public Integer getAge() {
        return age;
    }

    public void setAge(Integer age) {
        this.age = age;
    }

    public User() {

    }

    public User(String id, String username, String password, Integer age) {
        this.id = id;
        this.username = username;
        this.password = password;
        this.age = age;
    }
}

View Code

有这个警告之后，处理如下：

package com.sxd.entity;

import org.hibernate.annotations.GenericGenerator;

import javax.persistence.*;

@Entity
@GenericGenerator(name = "uuid2", strategy = "org.hibernate.id.UUIDGenerator" )
public class User {
    private String id;
    private String username;
    private String password;
    private Integer age;

    @Id
    @GeneratedValue(generator = "uuid2")
    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    @Column(nullable = false)
    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    @Column(nullable = false)
    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    @Column(nullable = false)
    public Integer getAge() {
        return age;
    }

    public void setAge(Integer age) {
        this.age = age;
    }

    public User() {

    }

    public User(String id, String username, String password, Integer age) {
        this.id = id;
        this.username = username;
        this.password = password;
        this.age = age;
    }
}

View Code

即可解决这个警告。