如何根据NOT条件连接2个实体

Question

我有两个表--例如TableA和TableB，其中有一些数据，如本文所定义的--如何基于not条件连接两个表的结果

现在我计划为两个表创建Hibernate实体，并且这些实体彼此不相关。

现在我想要得到基于NOT条件的结果，就像我前面提到的文章中给出的那样。

结果是使用SQL查询：

SELECT id, name, partNumber, Aid, Aname, Apart
FROM TableB AS t
CROSS JOIN (SELECT id AS Aid, name AS Aname, partNumber AS Apart
            FROM TableA AS a
            WHERE NOT EXISTS (SELECT 1
                              FROM TableB AS b
                              WHERE b.partNumber = a.partNumber)) AS c
ORDER BY id   

现在如何为这样的场景创建HQL查询或条件查询。我已经阅读了HQL&Criteria查询的Hibernate文档，但我不明白如何将此SQL查询转换为HQL和Criteria查询。你能帮我解决这个问题吗。

更新1：

根据弗拉德给出的答案，我没有得到正确的输出。

下面是我编写的代码：

List<Object[]> list = session.createQuery(
                "select a, b " + "from TableB b, TableA a "
                        + "where b.partNumber != a.partNumber "
                        + "ORDER BY b.id").list();

        for (Object[] objects : list) {
            for (Object object : objects) {
                System.out.println(object);
            }
        }

我得到以下输出：

A: id=2, name=a2, partNumber=20
B: id=5, name=b1, partNumber=10

A: id=3, name=a3, partNumber=30
B: id=5, name=b1, partNumber=10

A: id=4, name=a4, partNumber=40
B: id=5, name=b1, partNumber=10

A: id=1, name=a1, partNumber=10
B: id=6, name=b2, partNumber=20

A: id=3, name=a3, partNumber=30
B: id=6, name=b2, partNumber=20

A: id=4, name=a4, partNumber=40
B: id=6, name=b2, partNumber=20

A: id=1, name=a1, partNumber=10
B: id=7, name=b3, partNumber=60

A: id=2, name=a2, partNumber=20
B: id=7, name=b3, partNumber=60

A: id=3, name=a3, partNumber=30
B: id=7, name=b3, partNumber=60

A: id=4, name=a4, partNumber=40
B: id=7, name=b3, partNumber=60

A: id=1, name=a1, partNumber=10
B: id=8, name=b4, partNumber=70

A: id=2, name=a2, partNumber=20
B: id=8, name=b4, partNumber=70

A: id=3, name=a3, partNumber=30
B: id=8, name=b4, partNumber=70

A: id=4, name=a4, partNumber=40
B: id=8, name=b4, partNumber=70

从输出中，我获得了ID为1,2,3,4的TableA的记录&对于ID为TableBID'S=5,6,7,8。

但所需输出的TableA的ID应为3&4，而TableB的ID应为5、6、7、8。细节在我的另一篇文章中给出：如何基于not条件连接2个表的结果

Hibernate生成的查询是：

Hibernate: 
    /* select
        a,
        b 
    from
        TableB b,
        TableA a 
    where
        b.partNumber != a.partNumber 
    ORDER BY
        b.id */ 

select
    tablea1_.id as id1_0_0_,
    tableb0_.id as id1_1_1_,
    tablea1_.name as name2_0_0_,
    tablea1_.partNumber as partNumber3_0_0_,
    tableb0_.name as name2_1_1_,
    tableb0_.partNumber as partNumber3_1_1_ 
from
    TableB tableb0_ cross 
join
    TableA tablea1_ 
where
    tableb0_.partNumber<>tablea1_.partNumber 
order by
    tableb0_.id

更新2：

我现在尝试过的代码：

List<Object[]> list = session.createQuery("select b, a "
                + "from TableB b, TableA a "
                + "where not exists ( "
                + "select 1 "
                + "from TableB b1, TableA a1 "
                + "where "
                + "b1.partNumber = a1.partNumber and "
                + "b1.id = b.id and "
                + "a1.id = a.id " 
                + ") "
                + "order by b.id").list();
        for (Object[] objects : list) {
            for (Object object : objects) {
                System.out.println(object);
            }
        }

Hibernate: 

select
            tableb0_.id as id1_1_0_,
            tablea1_.id as id1_0_1_,
            tableb0_.name as name2_1_0_,
            tableb0_.partNumber as partNumb3_1_0_,
            tablea1_.name as name2_0_1_,
            tablea1_.partNumber as partNumb3_0_1_ 
        from
            TableB tableb0_ cross 
        join
            TableA tablea1_ 
        where
            not (exists (select
                1 
            from
                TableB tableb2_ cross 
            join
                TableA tablea3_ 
            where
                tableb2_.partNumber=tablea3_.partNumber 
                and tableb2_.id=tableb0_.id 
                and tablea3_.id=tablea1_.id)) 
        order by
            tableb0_.id

此查询得输出：

B: id=5, name=b1, partNumber=10
A: id=4, name=a4, partNumber=40
B: id=5, name=b1, partNumber=10
A: id=2, name=a2, partNumber=20
B: id=5, name=b1, partNumber=10
A: id=3, name=a3, partNumber=30
B: id=6, name=b2, partNumber=20
A: id=1, name=a1, partNumber=10
B: id=6, name=b2, partNumber=20
A: id=4, name=a4, partNumber=40
B: id=6, name=b2, partNumber=20
A: id=3, name=a3, partNumber=30
B: id=7, name=b3, partNumber=60
A: id=3, name=a3, partNumber=30
B: id=7, name=b3, partNumber=60
A: id=1, name=a1, partNumber=10
B: id=7, name=b3, partNumber=60
A: id=4, name=a4, partNumber=40
B: id=7, name=b3, partNumber=60
A: id=2, name=a2, partNumber=20
B: id=8, name=b4, partNumber=70
A: id=3, name=a3, partNumber=30
B: id=8, name=b4, partNumber=70
A: id=1, name=a1, partNumber=10
B: id=8, name=b4, partNumber=70
A: id=4, name=a4, partNumber=40
B: id=8, name=b4, partNumber=70
A: id=2, name=a2, partNumber=20

Answer 1

您需要使用theta样式的连接：

select b, a
from TableB b, TableA a 
where not exists (
    select 1
    from TableB b1, TableA a1
    where 
        b1.partNumber = a1.partNumber and
        b1.id = b.id and
        a1.id = a.id    
)   
order by b.id

也可以使用SQL查询来获取实体：

List result = session.createSQLQuery("SELECT b.*, c.* \n" +
        "FROM TableB b AS t\n" +
        "CROSS JOIN (SELECT id AS Aid, name AS Aname, partNumber AS Apart\n" +
        "            FROM TableA AS a\n" +
        "            WHERE NOT EXISTS (SELECT 1\n" +
        "                              FROM TableB AS b\n" +
        "                              WHERE b.partNumber = a.partNumber)) AS c\n" +
        "ORDER BY b.id ")
        .addEntity("b", B.class)
        .addEntity("a", A.class)
        .list();