[英]How to use countDistinct in Doctrine query builder (Symfony)
[英]How to get documents checking against a permissions junction table with query builder in symfony doctrine?
我很好奇如何使用 Symfony 中的查詢生成器構建 doctrine 查詢,它可以根據相當標准化的權限聯結表排除項目。
就我而言,
User
實體
Template
實體
userTemplatePermissionTable
在 userTemplatePermissions 實體中,我們有以下屬性:
userToCheck
, template
, read
, write
, delete
我想要的東西相當於:
return $this->createQueryBuilder('t')
[where count of t.userDocumentPermissions > 0
where canRead = true, canWrite = true, canDelete = true
AND userDocumentpermissions.user = :user]
這最終是一個相當簡單的左連接和排除檢查設置,我在我的TemplateRespository.php
中完成了以下操作
public function findUserAuthorized($user, $read = null, $write = null, $delete = null)
{
$query = $this->createQueryBuilder('t')
->leftJoin('t.userTemplatePermissions', 'utp')
->where('utp.userToCheck = :user')
->setParameter('user',$user);
if ($read !== null) {
$query->andWhere('utp.read = :read')
->setParameter('read',$read);
}
if ($write !== null) {
$query->andWhere('utp.write = :write')
->setParameter('write',$write);
}
if ($delete !== null) {
$query->andWhere('utp.delete = :delete')
->setParameter('delete',$delete);
}
return $query->getQuery()->getResult();
}
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