php - 向现有实体添加额外的 SELECT 表达式结果

Question

我有一个项目可以从多个方面对一家公司进行评级。下图可以帮助您了解我的数据库中的关系。当然，我已经从方案中删除了在我的问题上下文中不重要的字段。 目前，在我的 Company 实体的存储库类中，我正在使用 QueryBuilder 搜索公司并返回 \Doctrine\ORM\Tools\Pagination\Paginator 的实例。

下面的代码负责获取公司:

public function search(CompanySearchQuery $command): Paginator
{
    $qb = $this->createQueryBuilder('c')
        ->setFirstResult($command->getOffset())
        ->setMaxResults($command->getMaxResults())
        ->orderBy('c.name', 'ASC');

    ... plus extra "where" conditions

    return new Paginator($qb, true);
}

我想知道如何使用 QueryBuilder 和 Paginator 将平均分附加到每家公司。

这是允许我获取所需数据的 native SQL:

SELECT
  c.id,
  c.name,
  AVG(o2.score) as score
FROM
  company c
LEFT JOIN
  opinion o ON c.id = o.company_id
LEFT JOIN
  opinion_scope o2 ON o.id = o2.opinion_id
GROUP BY
  c.id

我的问题是: 是否可以将 averageScore 属性添加到 Company 类并将其映射到 QueryBuilder 结果？

我尝试重写我的 SQL 查询以使用现有代码:

$qb = $this->createQueryBuilder('c')
    ->select('c', 'AVG(s.score)')
    ->leftJoin('c.opinions', 'o')
    ->leftJoin('o.scopes', 's')
    ->groupBy('c.id')
    ->setFirstResult($command->getOffset())
    ->setMaxResults($command->getMaxResults())
    ->orderBy('c.name', 'ASC')
;

使用上面的代码，我得到如下数据库异常:

SQLSTATE[42000]: Syntax error or access violation: 1055 Expression #14 of 
SELECT list is not in GROUP BY clause and contains nonaggregated column 
'database.o1_.score' which is not functionally dependent on columns in GROUP BY
clause; this is incompatible with sql_mode=only_full_group_by").

为上述 QueryBuilder 执行的 SQL 查询是:

SELECT 
  DISTINCT id_8 
FROM 
  (
    SELECT 
      DISTINCT id_8, 
      name_9 
    FROM 
      (
        SELECT 
          c0_.updated_at AS updated_at_0, 
          c0_.description AS description_1, 
          c0_.website AS website_2, 
          c0_.email AS email_3, 
          c0_.phone AS phone_4, 
          c0_.street AS street_5, 
          c0_.postal_code AS postal_code_6, 
          c0_.city AS city_7, 
          c0_.id AS id_8, 
          c0_.name AS name_9, 
          c0_.slug AS slug_10, 
          c0_.created_at AS created_at_11, 
          AVG(o1_.score) AS sclr_12, 
          o1_.score AS score_13, 
          o1_.opinion_id AS opinion_id_14, 
          o1_.type_id AS type_id_15 
        FROM 
          company c0_ 
          LEFT JOIN opinion o2_ ON c0_.id = o2_.company_id 
          LEFT JOIN opinion_scope o1_ ON o2_.id = o1_.opinion_id 
        GROUP BY 
          c0_.id
      ) dctrn_result_inner 
    ORDER BY 
      name_9 ASC
  ) dctrn_result 
LIMIT 
  2 OFFSET 0

我不明白为什么 Doctrine 添加了以下片段:

o1_.score AS score_13, 
o1_.opinion_id AS opinion_id_14, 
o1_.type_id AS type_id_15 

Answer 1

问题似乎是您试图从“c”中选择每一列，这是执行两个 LEFT JOIN 的结果表。相反，您需要为公司表指定一个 from 子句:

$qb = $this->createQueryBuilder()
    ->select('c', 'AVG(s.score)')
    ->from('company', 'c')
    ->leftJoin('c.opinions', 'o')
    ->leftJoin('o.scopes', 's')
    ->groupBy('c.id')
    ->setFirstResult($command->getOffset())
    ->setMaxResults($command->getMaxResults())
    ->orderBy('c.name', 'ASC')
;

这将避免包含 o1.score 是您返回的列，这是错误引用的内容