Question

我有三个型号：位置＆gt;热点＆gt;帐户

在我的位置模型中，我有这个：

has_many :hotspots
has_many :accounts, :foreign_key => :calledstationidclean, :through => :hotspots

我正在尝试显示每个位置的帐户模型的一些摘要数据。

以下工作正常：

  scope :location_history, lambda { |id|
                               joins('INNER JOIN hotspots b ON locations.id = b.location_id INNER JOIN account ON b.mac = account.calledstationidclean')
                               .where(:id => id)
                               .select("count(locations.id) AS sessions_count, sum(account.acctoutputoctets) AS downloads, sum(account.acctinputoctets) AS uploads, count(distinct account.callingstationid) AS unique_user")
                            }

但我需要进行一些计算，我无法弄清楚如何做到这一点。

通常，在帐户模型中，我可以执行以下操作：

 def self.session_time
   Account.sum("acctsessiontime")
 end

 def self.average_session
   self.session_time / Account.count
 end

我如何在我的位置模型中做类似的事情 - 并且只显示我正在查看的特定位置的信息？我真的需要运行联接吗？

Answer 1

是的，你必须加入。

  scope :location_history,
        lambda { |id| joins(:hotspots, :accounts).where(:id => id).
                      select("sum(account.acctsessiontime) as acctsessiontime, count(locations.id) AS sessions_count, sum(account.acctoutputoctets) AS downloads, sum(account.acctinputoctets) AS uploads, count(distinct account.callingstationid) AS unique_user")
               }

顺便说一下，您可以使用一个组并将所有这些组合到位置的记录中。

  scope :with_history,
        lambda { |id| joins(:hotspots, :accounts).where(:id => id).
                      select("locations.*, sum(account.acctsessiontime) as acctsessiontime, count(locations.id) AS sessions_count, sum(account.acctoutputoctets) AS downloads, sum(account.acctinputoctets) AS uploads, count(distinct account.callingstationid) AS unique_user")
               }

这将返回包含所有位置字段的Location对象，以及您正在进行的聚合。

Rails3在三个方向上执行计算有很多通过关系

1 个答案: