hibernate – JPA CriteriaQuery包括2个子查询

2023年1月15日 336次阅读

我在使用JPA构建一个复杂的CriteriaQuery包括两个subquerys时遇到了问题.

在SQL中,查询看起来像这样：

SELECT * FROM photos p WHERE ((p.userID = 1) AND ((p.privacy = 0) 
OR (p.privacy = 1 AND EXISTS (SELECT * FROM friendships fs WHERE ((fs.userID = p.userID AND fs.userID2 = 2) OR (fs.userID = 2 AND fs.userID2 = p.userID))))
OR (p.privacy = 2 AND EXISTS (SELECT * FROM photo_privacy_users ppu WHERE (ppu.photoID = p.photoID AND ppu.userID = 2)))
));

澄清sql查询：
photos.privacy = 0表示PrivacyType.PUBLIC
photos.privacy = 1表示PrivacyType.PRIVATE
photos.privacy = 2表示PrivacyType.COSTUM

目前我收到以下错误：
org.hibernate.engine.jdbc.spi.SqlExceptionHelper – 操作数应包含1列
org.hibernate.exception.DataException：无法提取ResultSet

以下代码生成查询：

CriteriaBuilder cb = em.getCriteriaBuilder();

CriteriaQuery<Photo> cq = cb.createQuery(Photo.class);
Root<Photo> photo = cq.from(Photo.class);
cq.select(photo);

Subquery<Friendship> sq1 = cq.subquery(Friendship.class);
Root<Friendship> friendship = sq1.from(Friendship.class);
sq1.select(friendship);
sq1.where(cb.or(cb.and(cb.equal(friendship.get(Friendship_.userID), photo.get(Photo_.userID)), cb.equal(friendship.get(Friendship_.userID2), loggedinUserID)), cb.and(cb.equal(friendship.get(Friendship_.userID), loggedinUserID), cb.equal(friendship.get(Friendship_.userID2), photo.get(Photo_.userID)))));

Subquery<PhotoPrivacyUser> sq2 = cq.subquery(PhotoPrivacyUser.class);
Root<PhotoPrivacyUser> privacyUser = sq2.from(PhotoPrivacyUser.class);
sq2.select(privacyUser);
sq2.where(cb.and(cb.equal(privacyUser.get(PhotoPrivacyUser_.photoID), photo.get(Photo_.photoID)), cb.equal(privacyUser.get(PhotoPrivacyUser_.userID), loggedinUserID)));

Predicate predicate1 = cb.equal(photo.get(Photo_.userID), userID);
Predicate predicate2 = cb.equal(photo.get(Photo_.privacy), PrivacyType.PUBLIC);
Predicate predicate3 = cb.and(cb.equal(photo.get(Photo_.privacy), PrivacyType.PRIVATE), cb.exists(sq1));
Predicate predicate4 = cb.and(cb.equal(photo.get(Photo_.privacy), PrivacyType.COSTUM), cb.exists(sq2));

cq.where(cb.and(predicate1, cb.or(predicate2, predicate3, predicate4)));
return em.createQuery(cq).getResultList();

变量userID和loggedInUserID是Long类型的方法参数.

Metamodel Photo_：

@StaticMetamodel(Photo.class)
public abstract class Photo_ {
    public static volatile SingularAttribute<Photo, Long> photoID;
    public static volatile SingularAttribute<Photo, PrivacyType> privacy;
    public static volatile ListAttribute<Photo, PhotoPrivacyUser> photoPrivacyUsers;
    public static volatile SingularAttribute<Photo, String> name;
    public static volatile SingularAttribute<Photo, User> user;

}

Metamodel Friendship_：

@StaticMetamodel(Friendship.class)
public abstract class Friendship_ {

    public static volatile SingularAttribute<Friendship, User> user2;
    public static volatile SingularAttribute<Friendship, Date> createdAt;
    public static volatile SingularAttribute<Friendship, Long> userID2;
    public static volatile SingularAttribute<Friendship, Long> userID;
    public static volatile SingularAttribute<Friendship, User> user;

}

Metamodel PhotoPrivacyUser_：

@StaticMetamodel(PhotoPrivacyUser.class)
public abstract class PhotoPrivacyUser_ {

    public static volatile SingularAttribute<PhotoPrivacyUser, Date> createdAt;
    public static volatile SingularAttribute<PhotoPrivacyUser, Long> photoID;
    public static volatile SingularAttribute<PhotoPrivacyUser, Photo> photo;
    public static volatile SingularAttribute<PhotoPrivacyUser, Long> userID;
    public static volatile SingularAttribute<PhotoPrivacyUser, User> user;

}

元模型用户_：

@StaticMetamodel(User.class)
public abstract class User_ {
    public static volatile SingularAttribute<User, Long> userID;
    public static volatile ListAttribute<User, Photo> photos;
    public static volatile ListAttribute<User, Friendship> friends;
    public static volatile SingularAttribute<User, Date> createdAt;
    public static volatile ListAttribute<User, PhotoPrivacyUser> photoPrivacyUsers;
    public static volatile SingularAttribute<User, String> name;
    public static volatile SingularAttribute<User, String> email;
    public static volatile SingularAttribute<User, String> username;

}

如果我从查询中删除predicate3和predicate4,我的查询工作正常.
我不确定我是否应该使用subquerys或者为此加入.

最佳答案 我解决了我的问题.

我怀疑正确的方法是使用连接.

使用带存在的subquerys不能正常工作,因为exists返回的不仅仅是true或false.如果exists为true,则结果可以是多行.

以下代码生成具有三个连接的查询：

CriteriaBuilder cb = entityManager.getCriteriaBuilder();

CriteriaQuery<Photo> cq = cb.createQuery(Photo.class);
Root<Photo> photo = cq.from(Photo.class);

Join<Photo, User> userJoin = photo.join(Photo_.user, JoinType.LEFT);
Join<Photo, PhotoPrivacyUser> ppuJoin = photo.join(Photo_.photoPrivacyUsers, JoinType.LEFT);
Join<User, Friendship> fsJoin = userJoin.join(User_.friends, JoinType.LEFT);

cq.select(photo);

Predicate predicateUserID = cb.equal(photo.get(Photo_.userID), userID);
Predicate predicatePublic = cb.equal(photo.get(Photo_.privacy), PrivacyType.PUBLIC);

Predicate predicateFriendship = cb.or(cb.and(cb.equal(fsJoin.get(Friendship_.userID), photo.get(Photo_.userID)), cb.equal(fsJoin.get(Friendship_.userID2), loggedinUserID)), cb.equal(fsJoin.get(Friendship_.userID), loggedinUserID), cb.equal(fsJoin.get(Friendship_.userID2), photo.get(Photo_.userID)));
Predicate predicatePrivate = cb.and(cb.equal(photo.get(Photo_.privacy), PrivacyType.PRIVATE), predicateFriendship);

Predicate predicatePpu = cb.and(cb.equal(ppuJoin.get(PhotoPrivacyUser_.photoID), photo.get(Photo_.photoID)), cb.equal(ppuJoin.get(PhotoPrivacyUser_.userID), loggedinUserID));
Predicate predicateCostum = cb.and(cb.equal(photo.get(Photo_.privacy), PrivacyType.COSTUM), predicatePpu);

cq.where(cb.and(predicateUserID, cb.or(predicatePublic, predicatePrivate, predicateCostum)));

return entityManager.createQuery(cq).getResultList();

也许这会帮助有同样问题的人或者如何做出类似的事情. 🙂

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