有實體類user 和Info ，info對應的表中有user的主鍵作為外鍵，需求是查詢info可以得到對應的user,查詢user可以得到對應的info。package?com.zxy.demo.entity; import?com.fasterxml.jackson.annotation.JsonIgnoreProperties; import?com.fasterxml.jackson.annotation.JsonIgnoreType; import?javax.persistence.*; @Entity public?class?User?{ ????@Id ????@GeneratedValue ????private?Integer?id; ????private?Integer?age; ????@OneToOne(mappedBy?=?"user",cascade?=?CascadeType.ALL) ????private?Info?userInfo; ????public?Integer?getId()?{ ????????return?id; ????} ????public?void?setId(Integer?id)?{ ????????this.id?=?id; ????} ????public?Integer?getAge()?{ ????????return?age; ????} ????public?void?setAge(Integer?age)?{ ????????this.age?=?age; ????} ????public?Info?getUserInfo()?{ ????????return?userInfo; ????} ????public?void?setUserInfo(Info?userInfo)?{ ????????this.userInfo?=?userInfo; ????} }package?com.zxy.demo.entity; import?javax.persistence.*; @Entity public?class?Info?{ ????@Id ????@GeneratedValue ????private?Integer?id; ????@Column(nullable?=?false) ????private?String?idCard; ????@OneToOne(cascade?=?CascadeType.ALL) ????@JoinColumn(unique?=?true) ????private?User?user; ????public?Integer?getId()?{ ????????return?id; ????} ????public?void?setId(Integer?id)?{ ????????this.id?=?id; ????} ????public?String?getIdCard()?{ ????????return?idCard; ????} ????public?void?setIdCard(String?idCard)?{ ????????this.idCard?=?idCard; ????} ????public?User?getUser()?{ ????????return?user; ????} ????public?void?setUser(User?user)?{ ????????this.user?=?user; ????} ????@Override ????public?String?toString()?{ ????????return?"Info{"?+ ????????????????"id="?+?id?+ ????????????????",?idCard='"?+?idCard?+?'\''?+ ????????????????",?user="?+?user?+ ????????????????'}'; ????} }當調用jpa的findOne方法時，hibernate會注入user中的info,然后又一次注入info中的user，這樣很蠢。網上找到控制json解析的注解，這樣做還不如直接單項關聯，治標不治本。求一個解決方案！