原则：不能通过识别变量选择实体，而不选择至少一个根实体别名

问题描述

我在查询构建器中使用以下代码来选择分数值的平均值，以及该平均值所属的类别实体：

I'm using the following code in the query builder, to select an average of score values, and the category entity to which that average belongs:

$queryBuilder = $this->createQueryBuilder('s')
    ->resetDQLPart('select')
    ->select('AVG(s.score) as score, partial c.{reviewCategoryID} as cat')
    ->setParameter('status', ReviewStatusType::ACCEPTED)
    ->join('s.review', 'r')
    ->join('s.category', 'c')
    ->where('r.campsite = :campsite')
    ->andWhere('r.status = :status')
    ->setParameter('campsite', $campsite)
    ->groupBy('c.reviewCategoryID');

$ campsite 是一个实体，审查属于，而分数属于审查，分数有一个类别。

$campsite is an entity to which a review belongs, while scores belong to a review, and scores have a category.

但是当我尝试执行这个，我得到错误

But when I try to execute this, i get the error

Error: Cannot select entity through identification variables without choosing at least one root entity alias.

当我调试并检查根别名时，我看到's'被定义，应该是根实体（Score）。

When I debug and I check the root aliases, I see that 's' is defined, which is should be the root entity (Score).

任何想法可能是错的？

推荐答案

createQueryBuilder（）只能从匹配实体的存储库中调用参数。如果你不从这个存储库中调用它，你应该定义一个from方法。

createQueryBuilder() can only take a parameter when it is called from the repository of the matching entity. In case you do not call it from this repository you should define a from method.

->from('YourMappingSpace:Campsite', 's')

将参数传递给createQueryBuilder（）无论如何也是可以的。您可以随时手动定义它。该函数看起来像这样（只在实体存储库中）：

Passing a parameter to createQueryBuilder() is for conveniance anyway. You can always define it manually. The function looks like this (Only inside the entity repository):

public function createQueryBuilder($alias)
{
    return $this->_em->createQueryBuilder()
        ->select($alias)
        ->from($this->_entityName, $alias);
}

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