Question

对于一个项目，我需要以JSON格式提供大量不同的数据。所有信息都将在同一页面上使用，因此单个调用将导致最小的开销。所有信息都涉及相同的数据库对象，并且在页面上都是必需的。它基本上是一个特定类型中的一个或多个的对象数量的集合（这些类型都是布尔值），我们需要知道很多不同的变体。我使用下面的代码，但我的同事认为我把它放在JSON列表中的方式有点笨拙，代码可以有更好的性能。我怎么能改进这段代码？

public function getContactsStatisticsAction()
{
    $response = new Response();
    $json = array();
    $em = $this->getDoctrine()->getEntityManager();
    $cr = $em->getRepository('BlaCoreBundle:Company');

    $json['numberOfCompanies'] = $cr->numberOfCompanies();
    $json['numberOfAccounts'] = $cr->numberOfCompanies(array("typeAccount" => true));
    $json['numberOfCompetitors'] = $cr->numberOfCompanies(array("typeCompetitor" => true));
    $json['numberOfSuppliers'] = $cr->numberOfCompanies(array("typeSupplier" => true));
    $json['numberOfOthers'] = $cr->numberOfCompanies(array("typeOther" => true));
    $json['numberOfUnassigned'] = $cr->numberOfCompanies(array("typeAccount" => false, "typeCompetitor" => false,"typeSupplier" => false,"typeOther" => false));

    $json['numberOfJustAccounts'] = $cr->numberOfCompanies(array("typeAccount" => true, "typeCompetitor" => false, "typeSupplier" => false));
    $json['numberOfJustCompetitors'] = $cr->numberOfCompanies(array("typeAccount" => false, "typeCompetitor" => false, "typeSupplier" => false));
    $json['numberOfJustSuppliers'] = $cr->numberOfCompanies(array("typeAccount" => false, "typeCompetitor" => false, "typeSupplier" => false));

    $json['numberOfCompetitorAndAccounts'] = $cr->numberOfCompanies(array("typeAccount" => true, "typeCompetitor" => true, "typeSupplier" => false));
    $json['numberOfCompetitorAndSuppliers'] = $cr->numberOfCompanies(array("typeAccount" => false, "typeCompetitor" => true, "typeSupplier" => true));
    $json['numberOfSupplierAndAccounts'] = $cr->numberOfCompanies(array("typeAccount" => true, "typeCompetitor" => false, "typeSupplier" => true));
    $json['numberOfCompaniesAndAccountsAndSuppliers'] = $cr->numberOfCompanies(array("typeAccount" => true, "typeCompetitor" => true, "typeSupplier" => true));

    $response->setContent(json_encode($json));
    return $response;
}


public function numberOfCompanies($filters = array())
{
    $qb = $this->getEntityManager()->createQueryBuilder();
    $qb->select('count(c.id)');
    $qb->from('BlaCoreBundle:Company', 'c');
    $sizeFilters = count ($filters);
    $keys = array_keys($filters);
    if($sizeFilters >= 1){
        $qb->where('c.' . $keys[0] . ' = ' . (int) $filters[$keys[0]]);
    }
    for($i = 1; $i < $sizeFilters; $i++){
        $qb->andWhere('c.' . $keys[$i] . ' = ' . (int) $filters[$keys[$i]]);
    }
    return $qb->getQuery()->getSingleScalarResult();
}

Answer 1

你的同事是对的。您应该在单个查询中获得所有标量结果。您将以这种方式最小化连接数。

对于非Doctrine案例，this answer解决了该主题。

这样的用户也提出了这个有趣的问题here，但都没有回答。

实际上我认为没有办法用 QueryBuilder 或 DQL 来解决这种查询。此外，在 Doctrine2.2 的官方文档中，JOIN上没有SELECT的示例。

您可以尝试的内容类似于以下DQL查询：

return $this->getEntityManager()->createQuery(
  SELECT COUNT(c1) AS C1, COUNT(c2) AS C2, COUNT(c3) AS C3 FROM 
    BlaCoreBundle:Company c1, BlaCoreBundle:Company c2, BlaCoreBundle:Company c3
     WHERE c1.prop1 = 'xxx' AND c2.prop2 > '100' AND c3.prop3 LIKE '%XYZ%')               
  ->getResult();

当然WHERE子句是一般例子。此查询将返回一个一个大小的数组，其中C1，C2和C3作为您计算的值的键。当然，使用JOIN变得很困难，无论你需要什么，但你总是可以使用WHERE IN (SELECT...)和WHERE EXISTS (SELECT...)，例如

  SELECT COUNT(c1) AS C1, COUNT(c2) AS C2, COUNT(c3) AS C3 FROM 
    BlaCoreBundle:Company c1, BlaCoreBundle:Company c2, BlaCoreBundle:Company c3
     WHERE c1.prop1 = 'xxx' AND c2.prop2 > '100' AND c3.prop3 LIKE '%XYZ%'
 AND EXISTS (SELECT x FROM BlaCoreBundle:Entity x JOIN x.company comp WHERE x.prop = "valye" AND comp = c1)

使用Symfony 2.1和Doctrine 2对同一实体类型进行多次COUNT查询

1 个答案: