原则：用条件计算实体的项目

问题描述

我如何计算带有条件的实体的项目？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。我只想统计那些类型为employee的人员。我可以做这样的事情：

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，但并非最佳选择。是否有以下内容：？

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

那么，您可以使用来设置 COUNT 个查询：

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

假设 $ dm 是您的实体管理员。

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();

或者您也可以将其写在：

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，而不是对象，无法回忆起。如果是这样，只需将 COUNT（u）或-> count（'u'）更改为 COUNT（u.id）或-> count（'u.id'）或任何称为主键的字段。

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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