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[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 中的查询生成器构建 doctrine 查询,它可以根据相当标准化的权限联结表排除项目。
就我而言,
User
实体
Template
实体
userTemplatePermissionTable
在 userTemplatePermissions 实体中,我们有以下属性:
userToCheck
, template
, read
, write
, delete
我想要的东西相当于:
return $this->createQueryBuilder('t')
[where count of t.userDocumentPermissions > 0
where canRead = true, canWrite = true, canDelete = true
AND userDocumentpermissions.user = :user]
这最终是一个相当简单的左连接和排除检查设置,我在我的TemplateRespository.php
中完成了以下操作
public function findUserAuthorized($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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