学说:用条件计算实体的项目

2022-01-16 00:00:00 php doctrine doctrine-orm

在 Doctrine 中，我如何计算一个实体的物品?例如，我意识到我可以使用:

How can I count an entity's items with a condition in Doctrine? For example, I realize that I can use:

$usersCount = $dm->getRepository('User')->count();

但这只会计算所有用户.我想只计算那些具有员工类型的人.我可以这样做:

But that will only count all users. I would like to count only those that have type employee. I could do something like:

$users = $dm->getRepository('User')->findBy(array('type' => 'employee'));
$users = count($users);

这可行，但不是最佳的.是否有类似以下内容:?

That works but it's not optimal. Is there something like the following:? 

$usersCount = $dm->getRepository('User')->count()->where('type', 'employee');

推荐答案

好吧，你可以使用 QueryBuilder 设置 COUNT 查询:

Well, you could use the QueryBuilder to setup a COUNT query:

假设 $dm 是您的实体经理.

Presuming that $dm is your entity manager.

$qb = $dm->createQueryBuilder();

$qb->select($qb->expr()->count('u'))
   ->from('User', 'u')
   ->where('u.type = ?1')
   ->setParameter(1, 'employee');

$query = $qb->getQuery();

$usersCount = $query->getSingleScalarResult();

或者你可以把它写在 DQL:

Or you could just write it in DQL:

$query = $dm->createQuery("SELECT COUNT(u) FROM User u WHERE u.type = ?1");
$query->setParameter(1, 'employee');

$usersCount = $query->getSingleScalarResult();

计数可能需要在 id 字段上，而不是在对象上，无法回忆.如果是这样，只需将 COUNT(u) 或 ->count('u') 更改为 COUNT(u.id) 或 >->count('u.id') 或任何你的主键字段.

The counts might need to be on the id field, rather than the object, can't recall. If so just change the COUNT(u) or ->count('u') to COUNT(u.id) or ->count('u.id') or whatever your primary key field is called.

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