Hibernate只执行查询8次

我认为我的JUnit测试会比我用单词更好地解释这个问题！Hibernate只执行查询8次

@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); 
    } 
}

对于在控制台上的结果是：

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(); 
}

我做了另一个试验，与其他查询

@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); 
    } 
}

下面是结果：

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

所以，I'm使用Spring数据...可能是什么问题？

来源

2013-02-19 Marco Noronha

你是如何运行你的测试？我不能相信你原来的查询被命名为query8times是巧合。让我相信你可能有一个测试类缓存在某个地方。 – Perception 2013-02-19 13:09:31

为了确保，使它打印“Testing”+ n +“times”并在for循环中使用* n *。 – 2013-02-19 13:10:50

我怀疑你正在使用一个在更改'query8times'方法之前编译的类文件。 – 2013-02-19 13:32:15

你不应该只创建一个EntityManager吗？

EntityManager entityManager = entityManagerFactory.createEntityManager();

^行应该在executeSQL方法之外。

来源

2013-02-19 13:10:05 adarshr

是的，我只创建了一个entityManager，就像adarshr说的那样。但那不是问题。我忘了提交事务，关闭实体管理器...

这是最后的结果是：

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

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

    List<?> list = q.list(); 

    entityManager.getTransaction().commit(); 
    entityManager.close(); 

    return list; 
}

来源

2013-02-19 13:15:50

您应该将其作为评论或编辑原始问题发布，因为这不是答案。 – 2013-02-19 13:54:58

Hibernate只执行查询8次

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