我有2个实体,我们之间有很多关系。我想在关联表上写一个查询。
Here are the two entities. Craftsmen and Skill.
I want to write a query which selects all Craftsmen with a given skill.
@Entity
@Table(name = "craftsman")
public class Craftsman {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@ManyToMany
@JoinTable(name = "craftman_skill", joinColumns = @JoinColumn(name = "craftman_id", referencedColumnName = "id"),
inverseJoinColumns = @JoinColumn(name = "skill_id", referencedColumnName = "id"))
private List<Skill> skillList;
}
@Entity
@Table(name = "skill")
public class Skill {
@ManyToMany(mappedBy = "skillList")
@JsonBackReference
private List<Craftsman> craftmanList;
}
以下是我的尝试:
@Query("SELECT c FROM Craftsman c JOIN c.skillList sl WHERE c.skillList.skill_id = :skillId")
public List<Craftsman> getCraftsmanBySkill(@Param("skillId") Long skillId);
以下是在MySql中有效的查询: 这让所有拥有id 1技能的工匠都得到了:
select c.name from craftsman c, craftman_skill cs
where cs.craftman_id = c.id
and cs.skill_id = 1;
以下是我得到的错误:
Caused by: java.lang.IllegalArgumentException: org.hibernate.QueryException: could not resolve property: skill_id of: com.craftsmen.crafts.persistence.Skill [SELECT c FROM com.craftsmen.crafts.persistence.Craftsman c JOIN c.skillList sl WHERE c.skillList.skill_id = :skillId ]
Caused by: org.hibernate.QueryException: could not resolve property: skill_id of: com.craftsmen.crafts.persistence.Skill [SELECT c FROM com.craftsmen.crafts.persistence.Craftsman c JOIN c.skillList sl WHERE c.skillList.skill_id = :skillId ]
答案 0 :(得分:2)
我想问题是,您的Skill
实体确实没有skill_id,或者更好,id
。
您是否曾尝试向Skill
实体提供id
@Entity
@Table(name = "skill")
public class Skill {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@ManyToMany(mappedBy = "skillList")
@JsonBackReference
private List<Craftsman> craftmanList;
}
在您的Craftsman
实体
@Entity
@Table(name = "craftsman")
public class Craftsman {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@ManyToMany
@JoinTable(name = "craftman_skill", joinColumns = @JoinColumn(name = "craftman_id", referencedColumnName = "id"),
inverseJoinColumns = @JoinColumn(name = "skill_id", referencedColumnName = "id"))
private List<Skill> skillList;
}
然后将Query
更改为
SELECT c FROM Craftsman c JOIN c.skillList sl WHERE sl.id = :skillId
这应该可以解决问题。
答案 1 :(得分:0)
此外,您可以考虑在此处使用JpaRepository的选项:
public interface CraftsmanRepository extends JpaRepository<Craftsman, String> {
public List<Craftsman> getBySkillListIn(@NotEmpty List<Skill> skillList);
}
这意味着,Jpa,通过写入方法名称已经将查询映射到您的SQL查询(如果没有,请尝试)
在方法结束时,“在”意味着它应该检查不是一个值而是多个值
在SQL In
中也可以工作并执行以下操作:
where <condition> And <>condition And <condition> And...