java-多列上的JPA连接

我有实体User,Organization和GrantedRole.它们之间的关系是,GrantedRole定义了用户在组织中的角色.这三个实体都从其超类继承字段“ id”(用@Id注释).组织没有对GrantedRole或User的引用,这些关系是单向的.

@Entity
@Table(name = "role_assignments", uniqueConstraints = @UniqueConstraint(columnNames = {
    "user_id", "organization_id" }))
public class GrantedRole extends DomainEntity implements GrantedAuthority {

    public static enum Role {
        ROLE_MODM_USER, ROLE_MODM_ORGADMIN, ROLE_MODM_ADMIN
    }

    private User user;
    private Role role;
    private Organization organization;

    public static enum Role {
        ROLE_MODM_USER, ROLE_MODM_ORGADMIN, ROLE_MODM_ADMIN
    }

    @ManyToOne
    @NotNull
    public User getUser() {
        return user;
    }

    @NotNull
    @Enumerated(EnumType.STRING)
    public Role getRole() {
        return role;
    }

    @ManyToOne
    public Organization getOrganization() {
        return organization;
    }

    // Some setters
}

@Entity
@Table(name = "modmUsers")
public class User extends DomainEntity implements UserDetails {
    // Some other fields

    private Set<GrantedRole> roles = new HashSet<GrantedRole>();
    private Set<Organization> organizations = new HashSet<Organization>();

    @OneToMany(fetch = FetchType.EAGER, mappedBy = "user")
    public Set<GrantedRole> getRoles() {
        return roles;
    }

    @ManyToMany(fetch = FetchType.EAGER)
    public Set<Organization> getOrganizations() {
        return organizations;
    }

    // Some setters
}

现在,我想使用JPA条件来查找用户具有ROLE_MODM_ORGADMIN角色的组织.但是,生成的查询将连接在单个列上,并导致重复GrantedRole中的行,这是因为User.id在GrantedRole中不是唯一的,而Organization.id也不是唯一的.两者的结合是独特的.

我找到组织的代码：

public List<Organization> findOrganizationsByOrgAdminUser(String 
    CriteriaBuilder cb = entityManager.
    CriteriaQuery<Organization> query = cb.createQuery(Organization.
    Root<User> root = query.from(User.class);
    SetJoin<User, Organization> joinOrg = root.joinSet("organizations");
    SetJoin<User, GrantedRole> joinRoles = root.joinSet("roles");

    Predicate p1 = cb.equal(root.get("id"), userId);
    Predicate p2 = cb.equal(joinRoles.get("role"), Role.ROLE_MODM_ORGADMIN);

    query.select(joinOrg);
    query.where(cb.and(p1, p2));

    return entityManager.createQuery(query).getResultList();
}

生成的查询是：

SELECT
    orgs.* 
FROM
    modmUsers users
INNER JOIN
    modmUsers_organizations u_o 
        ON users.id=u_o.modmUsers_id 
INNER JOIN
    organizations orgs
        ON u_o.organization_id=orgs.id 
INNER JOIN
    role_assignments roles 
        ON users.id=roles.user_id 
WHERE
    users.id=? 
    and roles.role=?

我想要的查询是：

SELECT
    orgs.* 
FROM
    modmUsers users
INNER JOIN
    modmUsers_organizations u_o 
        ON users.id=u_o.modmUsers_id 
INNER JOIN
    organizations orgs
        ON u_o.organization_id=orgs.id 
INNER JOIN
    role_assignments roles 
        ON users.id=roles.user_id 
        AND orgs.id=roles.organization_id //how to do this???
WHERE
    users.id=? 
    and roles.role=?

解决方法:

好的,所以我的同事帮助了我,并导致我执行以下查询：

SELECT
    orgs.* 
FROM
    modmUsers users
INNER JOIN
    modmUsers_organizations u_o 
        ON users.id=u_o.modmUsers_id 
INNER JOIN
    organizations orgs
        ON u_o.organization_id=orgs.id 
INNER JOIN
    role_assignments roles 
        ON users.id=roles.user_id 
WHERE
    users.id=? 
    AND roles.role=?
    AND orgs.id=roles.organization_id

这是我的代码的简单扩展：

public List<Organization> findOrganizationsByOrgAdminUser(String 
    CriteriaBuilder cb = entityManager.
    CriteriaQuery<Organization> query = cb.createQuery(Organization.
    Root<User> root = query.from(User.class);
    SetJoin<User, Organization> joinOrg = root.joinSet("organizations");
    SetJoin<User, GrantedRole> joinRoles = root.joinSet("roles");

    Predicate p1 = cb.equal(root.get("id"), userId);
    Predicate p2 = cb.equal(joinRoles.get("role"), Role.ROLE_MODM_ORGADMIN);
    Predicate p3 = cb.equal(joinOrg.get("id"), joinRoles.get("organization"));

    query.select(joinOrg);
    query.where(cb.and(p1, p2, p3));

    return entityManager.createQuery(query).getResultList();
}

标签：jpa,inner-join,java
来源： https://codeday.me/bug/20191030/1970056.html