如何基于NOT条件加入2个实体

Question

我有2个表 - 例如TableA和TableB,其中包含一些数据,如本文所定义 - 如何基于非条件连接2个表的结果

现在我计划为这两个表创建Hibernate实体,并且实体彼此不相关.

现在我想根据我之前提到的帖子中给出的NOT条件得到结果.

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,如何为此类场景创建HQL查询或条件查询.我已经浏览了Hibernate文档中的HQL和Criteria查询,但我无法理解如何将此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.

这是我写的代码:

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);
            }
        }

我得到了以下输出:

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

从输出中我得到TableA了id 1,2,3,4为= &的TableBid = s的记录5,6,7,8.

但所需的输出应该有ID的TableA作为3&4和TableBID的作为5,6,7,8.详细信息在我的另一篇文章中给出:如何根据非条件加入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生成的查询:

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

输出此查询:

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

或者您也可以使用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();