先贴上报错的图：

 报错出现的原因是，我要查询系统用户，使用jpa的findbyid查询，返回结果就会报错，系统用户的实体类是这样写的：

@Entity
@Data
@Table(name = "gspuser")
public class EsgspuserEntity {
    @Id
    @Column(name = "id",insertable=false,updatable=false)
    private String id;
    @Column(name = "code",insertable=false,updatable=false)
    private String code;
    @Column(name = "firstname_chs",insertable=false,updatable=false)
    private String firstnameChs;
    @Column(name = "firstname_cht",insertable=false,updatable=false)
    private String firstnameCht;
    @Column(name = "firstname_en",insertable=false,updatable=false)
    private String firstnameEn;
    @Column(name = "firstname_es",insertable=false,updatable=false)
    private String firstnameEs;
    @Column(name = "firstname_pt",insertable=false,updatable=false)
    private String firstnamePt;
    @Column(name = "lastname_chs",insertable=false,updatable=false)
    private String lastnameChs;
    @Column(name = "lastname_cht",insertable=false,updatable=false)
    private String lastnameCht;
    @Column(name = "lastname_en",insertable=false,updatable=false)
    private String lastnameEn;
    @Column(name = "lastname_es",insertable=false,updatable=false)
    private String lastnameEs;
    @Column(name = "lastname_pt",insertable=false,updatable=false)
    private String lastnamePt;
    @Column(name = "middlename_chs",insertable=false,updatable=false)
    private String middlenameChs;
    @Column(name = "middlename_cht",insertable=false,updatable=false)
    private String middlenameCht;
    @Column(name = "middlename_en",insertable=false,updatable=false)
    private String middlenameEn;
    @Column(name = "middlename_es",insertable=false,updatable=false)
    private String middlenameEs;
    @Column(name = "middlename_pt",insertable=false,updatable=false)
    private String middlenamePt;
    @Column(name = "name_chs",insertable=false,updatable=false)
    private String nameChs;
    @Column(name = "name_cht",insertable=false,updatable=false)
    private String nameCht;
    @Column(name = "name_en",insertable=false,updatable=false)
    private String nameEn;
    @Column(name = "name_es",insertable=false,updatable=false)
    private String nameEs;
    @Column(name = "name_pt",insertable=false,updatable=false)
    private String namePt;
    @Column(name = "sysorgid",insertable=false,updatable=false)
    private String sysorgid;
    @Column(name = "tenantid",insertable=false,updatable=false)
    private Integer tenantid;
    private String sex;
    @OneToMany
    @JoinColumn(name = "userid")
    private List<EsgspuserextendEntity> esgspuserextendEntityList;
}
可以看到在最后一个属性上，有@OneToMany的注解，根据在网上查询的结果，是查询懒加载导致的，这个时候需要在这个注解加上
@OneToMany(mappedBy="user",fetch=FetchType.EAGER)
我修改之后，程序还是报错，报错内容是和事务回滚有关系，我没有继续解决这个错误，又回去思考怎么解决原来的问题
因为我只需要系统用户的名字，所以我只要返回名字这个值即可，在修改了程序之后，果然没有继续报错。
此文只是为了记录工作中的问题，后续继续补充报错的原因和解决办法

SpringBoot通过jpa查询返回报错failed to lazily initialize a collection of role