Question

我试图在JPA Criteria Query中创建以下查询：

SELECT V.* FROM VENDA V
WHERE 
( SELECT COUNT(D.ID) FROM VENDADETALHE D WHERE D.ID IN
( SELECT CVD.DETALHES_ID FROM VENDA_VENDADETALHE CVD WHERE CVD.VENDA_ID = V.ID ) 
AND ( SELECT I.TIPO FROM ITEM I WHERE I.ID=D.IDITEM ) <> 'SERVICO' ) = 0
AND
( SELECT COUNT(D.ID) FROM VENDADETALHE D WHERE D.ID IN
( SELECT CVD.DETALHES_ID FROM VENDA_VENDADETALHE CVD WHERE CVD.VENDA_ID = V.ID ) 
AND ( SELECT I.TIPO FROM ITEM I WHERE I.ID=D.IDITEM ) = 'SERVICO' ) > 0

我有以下类结构

public class Venda {
  /* other class attributes ( id & etc... ) */
  @OneToMany( orphanRemoval=true, cascade = CascadeType.ALL, fetch =FetchType.LAZY )
  @JoinTable
  private List<VendaDetalhe> detalhes;
}

public class VendaDetalhe{
  /* other class attributes */
  @ManyToOne
  @JoinColumn( name = "iditem", referencedColumnName = "id")
  private Item item;
}

public class Item{
  /* other class attributes */
  @Enumerated( EnumType.STRING )
  private ETipo tipo;
}
public enum ETipo{
  PRODUTO,
  SERVICO;
}

以下代码查询：

CriteriaBuilder builder = em.getCriteriaBuilder();
CriteriaQuery<Venda> qry = builder.createQuery( Venda.class );
Root<Venda> root = qry.from( Venda.class );
Join<Venda,VendaDetalhe> join = root.join( "detalhes", JoinType.INNER );
List<Predicate> p = new ArrayList<>();
p.add( builder.greatherThan( builder.count( builder.equal( join.get( "item" ).get("tipo"), ETipo.SERVICO ) ), 0L ) );
p.add( builder.equal( builder.count( builder.equal( join.get("item" ).get("tipo"),ETipo.PRODUTO) ), 0L) );
qry.where( p.toArray( new Predicate[ p.size() ] );
em.createQuery( qry ).getResultList();

但这会生成QuerySyntaxException，表示预计会CLOSE，但会找到=。我的CriteriaQuery语法是否正确？我在互联网上搜索了在WHERE子句中使用count，但找不到任何内容。

异常消息如下：

org.hibernate.hql.internal.ast.QuerySyntaxException: expecting CLOSE, found '=' near line 1, 
column 249 
[select generatedAlias0 from
 br.com.koinonia.habil.model.user.movimentacoes.compraevenda.Venda 
as generatedAlias0 inner join generatedAlias0.detalhes as generatedAlias1 where 
( generatedAlias0.empresa=:param0 ) and ( count(generatedAlias1.item.tipo=:param1)>0L ) 
and ( count(generatedAlias1.item.tipo=:param2)=0L )]

Answer 1

Don't run a count query when you could run an exists query。即使您保持在SQL语言（或任何转换为SQL的语言，如JPQL）中，您的优化器也可能无法将COUNT(*) == 0转换为NOT EXISTS()并且COUNT(*) > 0到EXISTS()。想象一下，其中一项计数结果为100万。您真的需要数据库来确定完全计数值吗？或者，只要数据库知道是否存在（或没有）给定的行，数据库是否可以停止？

您的原始查询可以重写为：

SELECT V.* 
FROM VENDA V
WHERE NOT EXISTS ( 
  SELECT 1 
  FROM VENDADETALHE D 
  WHERE D.ID IN ( 
    SELECT CVD.DETALHES_ID 
    FROM VENDA_VENDADETALHE CVD 
    WHERE CVD.VENDA_ID = V.ID 
  ) 
  AND ( 
    SELECT I.TIPO FROM ITEM I WHERE I.ID=D.IDITEM 
  ) <> 'SERVICO' 
)
AND EXISTS ( 
  SELECT 1 
  FROM VENDADETALHE D 
  WHERE D.ID IN ( 
    SELECT CVD.DETALHES_ID 
    FROM VENDA_VENDADETALHE CVD 
    WHERE CVD.VENDA_ID = V.ID 
  ) 
  AND ( 
    SELECT I.TIPO FROM ITEM I WHERE I.ID=D.IDITEM 
  ) = 'SERVICO' 
)

当然，与'SERVICO'进行比较的相关子查询可以进一步转换为内部联接，但我不确定这是否会造成麻烦。

也许，您现在可以更轻松地使用JPQL编写，但为什么不通过EntityManager.createNativeQuery(String, Class)运行SQL查询。由于您正在预测一个实体（V.*），这样就可以了。

Answer 2

Hibernate中的

count QL对函数内部的内容显然非常挑剔，不能接受嵌套查询。所以这一行：

builder.count( builder.equal( join.get( "item" ).get("tipo"), ETipo.SERVICO ) )

需要稍微分开，以便equal比较在其他地方，count只在一个查询中调用。

JPA标准 - 依靠WHERE条款

2 个答案: