java - Hibernate只执行8次查询

Question

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已关闭 10 年前。

我认为我的 JUnit 测试将比我用文字更好地解释这一点!

    @Test
public void query8times(){

    for(int i=0; i<15; i++){
        ProspectoRadarQueryBuilder prqb = new ProspectoRadarQueryBuilder("jardeu");
        List<Object[]> prospectosNotas = (List<Object[]>) genericFilterDao.executeSQL(prqb.buildQuery());   
        System.out.println("------------------------------------- "+i);
    }
}

控制台上的结果是:

Hibernate:   SELECT * FROM   (   select         p.id,           comparestrings('jardeu', pc.valor) as nota   from           com_prospecto p         inner join com_prospecto_campo pc ON (p.id = pc.id_prospecto)           inner join com_campo c ON (pc.id_campo = c.id AND c.flag_nome = true)    ) as subQuery   where          nota is not null            AND         nota > 0.35  order by           nota desc;
------------------------------------- 0
Hibernate:   SELECT * FROM   (   select         p.id,           comparestrings('jardeu', pc.valor) as nota   from           com_prospecto p         inner join com_prospecto_campo pc ON (p.id = pc.id_prospecto)           inner join com_campo c ON (pc.id_campo = c.id AND c.flag_nome = true)    ) as subQuery   where          nota is not null            AND         nota > 0.35  order by           nota desc;
------------------------------------- 1
Hibernate:   SELECT * FROM   (   select         p.id,           comparestrings('jardeu', pc.valor) as nota   from           com_prospecto p         inner join com_prospecto_campo pc ON (p.id = pc.id_prospecto)           inner join com_campo c ON (pc.id_campo = c.id AND c.flag_nome = true)    ) as subQuery   where          nota is not null            AND         nota > 0.35  order by           nota desc;
------------------------------------- 2
Hibernate:   SELECT * FROM   (   select         p.id,           comparestrings('jardeu', pc.valor) as nota   from           com_prospecto p         inner join com_prospecto_campo pc ON (p.id = pc.id_prospecto)           inner join com_campo c ON (pc.id_campo = c.id AND c.flag_nome = true)    ) as subQuery   where          nota is not null            AND         nota > 0.35  order by           nota desc;
------------------------------------- 3
Hibernate:   SELECT * FROM   (   select         p.id,           comparestrings('jardeu', pc.valor) as nota   from           com_prospecto p         inner join com_prospecto_campo pc ON (p.id = pc.id_prospecto)           inner join com_campo c ON (pc.id_campo = c.id AND c.flag_nome = true)    ) as subQuery   where          nota is not null            AND         nota > 0.35  order by           nota desc;
------------------------------------- 4
Hibernate:   SELECT * FROM   (   select         p.id,           comparestrings('jardeu', pc.valor) as nota   from           com_prospecto p         inner join com_prospecto_campo pc ON (p.id = pc.id_prospecto)           inner join com_campo c ON (pc.id_campo = c.id AND c.flag_nome = true)    ) as subQuery   where          nota is not null            AND         nota > 0.35  order by           nota desc;
------------------------------------- 5
Hibernate:   SELECT * FROM   (   select         p.id,           comparestrings('jardeu', pc.valor) as nota   from           com_prospecto p         inner join com_prospecto_campo pc ON (p.id = pc.id_prospecto)           inner join com_campo c ON (pc.id_campo = c.id AND c.flag_nome = true)    ) as subQuery   where          nota is not null            AND         nota > 0.35  order by           nota desc;
------------------------------------- 6
Hibernate:   SELECT * FROM   (   select         p.id,           comparestrings('jardeu', pc.valor) as nota   from           com_prospecto p         inner join com_prospecto_campo pc ON (p.id = pc.id_prospecto)           inner join com_campo c ON (pc.id_campo = c.id AND c.flag_nome = true)    ) as subQuery   where          nota is not null            AND         nota > 0.35  order by           nota desc;
------------------------------------- 7

现在让我们看看源代码!

public List<?> executeSQL(String sql) {
    EntityManager entityManager = entityManagerFactory.createEntityManager();
    entityManager.getTransaction().begin();

    Session hibernateSession = entityManager.unwrap(Session.class);
    Query q = hibernateSession.createSQLQuery(sql);

    return q.list();
}

我用另一个查询做了另一个测试

@Test
public void anotherQuery(){

    for(int i=0; i<15; i++){
        List<Object[]> prospectosNotas = (List<Object[]>) genericFilterDao.executeSQL("select * from com_campo");   
        System.out.println("------------------------------------- "+i);
    }
}

结果如下:

Hibernate: select * from com_campo
------------------------------------- 0
Hibernate: select * from com_campo
------------------------------------- 1
Hibernate: select * from com_campo
------------------------------------- 2
Hibernate: select * from com_campo
------------------------------------- 3
Hibernate: select * from com_campo
------------------------------------- 4
Hibernate: select * from com_campo
------------------------------------- 5
Hibernate: select * from com_campo
------------------------------------- 6
Hibernate: select * from com_campo
------------------------------------- 7

那么，我正在使用 Spring Data...可能有什么问题？

Answer 1

您不应该只创建一个 EntityManager 吗？

EntityManager entityManager = entityManagerFactory.createEntityManager();

^ 行应该位于executeSQL 方法之外。