当我传递自定义的连接查询时,我遇到了有问题的捆绑问题。 我的主要目的是减少在20个结果分页时运行的查询,它执行超过30个查询。
但是使用带有连接的预定义查询,其他各种表“正常”减少为2个查询,当然,节省的性能不会受到伤害!
问题出现在KnpPaginatorBundle上,似乎使用我在30个查询中运行相同的查询。
控制器
public function indexAction(Request $request)
{
$em = $this->getDoctrine()->getManager();
$form = $this->createForm(new SoundFilterType());
if (!is_null($response = $this->saveFilter($form, 'sound', 'sound'))) {
return $response;
}
$qb = $em->getRepository('AcmeSoundBundle:Soundtrack')->findAllJoin();
$paginator = $this->filter($form, $qb, 'sound');
return array(
'form' => $form->createView(),
'paginator' => $paginator,
);
}
/**
* @param QueryBuilder $qb
* @param string $name
*/
protected function addQueryBuilderSort(QueryBuilder $qb, $name)
{
$alias = current($qb->getDQLPart('from'))->getAlias();
if (is_array($order = $this->getOrder($name))) {
$qb->orderBy($alias . '.' . $order['field'], $order['type']);
}
}
/**
* Filter form
*
* @param FormInterface $form
* @param QueryBuilder $qb
* @return SlidingPagination
*/
protected function filter(FormInterface $form, QueryBuilder $qb, $name)
{
if (!is_null($values = $this->getFilter($name))) {
if ($form->bind($values)->isValid()) {
$this->get('lexik_form_filter.query_builder_updater')->addFilterConditions($form, $qb);
}
}
$count = $this->getDoctrine()->getManager()
->createQuery('SELECT COUNT(c) FROM AcmeSoundBundle:Sound c')
->getSingleScalarResult();
// possible sorting
$this->addQueryBuilderSort($qb, $name);
return $this->get('knp_paginator')->paginate($qb->getQuery()->setHint('knp_paginator.count', $count), $this->getRequest()->query->get('page', 1), 20, array('distinct' => false));
}
查询
public function findAllJoin()
{
$qb = $this->createQueryBuilder('q');
$query = $qb->select('u')
->from('AcmeSoundBundle:Sound', 'u')
->innerJoin('u.artists', 'a')
->leftJoin('u.versions', 'v')
->leftJoin('v.instruments', 'i')
->addOrderBy('u.dataIns', 'ASC');
}
这是一个要点https://gist.github.com/Lughino/6358896
我不明白为什么它不使用我的查询..
有什么想法吗?
对于这个问题,我发现错误:
public function findAllJoin()
{
$qb = $this->createQueryBuilder('q');
$query = $qb->select('u, a, v, i')
->from('AcmeSoundBundle:Sound', 'u')
->innerJoin('u.artists', 'a')
->leftJoin('u.versions', 'v')
->leftJoin('v.instruments', 'i')
->addOrderBy('u.dataIns', 'ASC');
}
代替:
public function findAllJoin()
{
$qb = $this->createQueryBuilder('q');
$query = $qb->select('u')
->from('AcmeSoundBundle:Sound', 'u')
->innerJoin('u.artists', 'a')
->leftJoin('u.versions', 'v')
->leftJoin('v.instruments', 'i')
->addOrderBy('u.dataIns', 'ASC');
}
现在的问题是,如果我选择一个列进行排序,我只显示一个结果,即使我排序失败,实际上结果也没有排序。
我有什么遗失的吗?
解决! 如果我设置我的两个从实例化,并为此搞砸了! 这样:
public function findAllJoin()
{
$qb = $this->createQueryBuilder('u');
$query = $qb->select('u, a, v, i')
->innerJoin('u.artists', 'a')
->leftJoin('u.versions', 'v')
->leftJoin('v.instruments', 'i')
->addOrderBy('u.dataIns', 'ASC');
}
问题解决了!