Question

我有一个名为Tag的实体，它有3个ManyToMany关系的节目，文章和新闻。

AppBundle\Entity\Tag:
    type: entity
    table: tag
    ...
  manyToMany:
      news:
          orderBy: { 'posted': 'DESC' }
          targetEntity: News
          inversedBy: tags
          joinTable:
            name: news_tag
            joinColumns:
              tag_id:
                referencedColumnName: id
            inverseJoinColumns:
              news_id:
                referencedColumnName: id
      articles:
          orderBy: { 'posted': 'DESC' }
          targetEntity: Article
          inversedBy: tags
          joinTable:
            name: article_tag
            joinColumns:
              tag_id:
                referencedColumnName: id
            inverseJoinColumns:
              article_id:
                referencedColumnName: id
      fests:
          orderBy: { 'when_starts': 'DESC', 'when_ends': 'DESC' }
          targetEntity: Fest
          inversedBy: tags
          joinTable:
            name: fest_tag
            joinColumns:
              tag_id:
                referencedColumnName: id
            inverseJoinColumns:
              fest_id:
                referencedColumnName: id

现在我想编写一个createQueryBuilder，按计数“新闻”排序+计数“文章”+计数“fests”，仅用于查找“TOP TAGS”。我找到的解决方案只适用于一个很多关系。

有没有办法做到这一点？

编辑：我有什么：

    $qb->select(array(
        't.id', 
        't.name', 
        'COUNT(f) as festcount',
        'COUNT(n) as newscount',
        'COUNT(a) as articlescount',
        '(COUNT(f) + COUNT(n) + COUNT(a)) as totalcount'
        ))
    ->from('AppBundle:Tag', 't')
    ->leftJoin('t.fests', 'f')
    ->leftJoin('t.articles', 'a')
    ->leftJoin('t.news', 'n')
    ->groupBy('t.id')
    ->orderBy('totalcount', 'DESC');

但它给出了假结果。当一些标签有3个节目，1个文章和0个新闻时，结果是festcount = 3，newscount = 0，articlescount = 3和totalcount = 6，它应该是4。

Answer 1

试试这段代码：

$qb->select(array(
        't.id', 
        't.name', 
        'COUNT(DISTINCT f.id) as festcount',
        'COUNT(DISTINCT n.id) as newscount',
        'COUNT(DISTINCT a.id) as articlescount',
        '(festcount + newscount + articlescount) as totalcount'
        ))
    ->from('AppBundle:Tag', 't')
    ->leftJoin('t.fests', 'f')
    ->leftJoin('t.articles', 'a')
    ->leftJoin('t.news', 'n')
    ->groupBy('t.id')
    ->orderBy('totalcount', 'DESC');

如果您需要进一步的帮助，请提供这些表的转储。

Symfony3 querybuilder命令数不胜数

1 个答案: