轨道/阿雷尔：选择所有记录作为一个ActiveRecord ::关系 [英] Rails/Arel: Selecting all records as an ActiveRecord::Relation

问题描述

使用阿雷尔Rails中 - 我正在寻找创建的ActiveRecord ::关联，有效地导致了 SELECT * FROM表的方式，我可以进一步处理。

Using Arel in Rails - I'm looking for a way of creating an ActiveRecord::Relation that effectively results in SELECT * FROM table, which I can still manipulate further.

例如，我有一个模型，该模型分成多个类别，我在下面的方式返回计数这些：

For example, I have a model that's split up into multiple categories, and I return counts for these in the following manner:

relation = Model.where(:archived => false) # all non-archived records
record_counts = {
  :total => relation.count,
  :for_sale => relation.where(:for_sale => true).count
  :on_auction => relation.where(:on_auction => true).count
}

这工作得很好，并具有发射了 COUNT 查询到MySQL，而不是实际选择记录本身的优势。

This works fine, and has the advantage of firing off COUNT queries to MySQL, rather than actually selecting the records themselves.

不过，我现在需要包括在计数存档记录，但关系= Model.all 导致一个阵列，我正在寻找一个的ActiveRecord ::关联。

However, I now need to include archived records in the counts, but relation = Model.all results in an Array, and I'm looking for an ActiveRecord::Relation.

我能想到这样做的唯一方法是 model.where（model.arel_table [：ID] .not_eq（无）），它的工作原理，但略微显得荒谬的。

The only way I can think of doing this is model.where(model.arel_table[:id].not_eq(nil)), which works, but seems slightly absurd.

任何人都可以阐明这个任何光线？

Can anyone shed any light on this?

推荐答案

有关Rails的4.1及以上版本： Model.all 返回的关系（它previously做没有）

For Rails 4.1 and above: Model.all returns a relation (where it previously did not)

有关Rails的4.0： Model.where（无）

For Rails 4.0: Model.where(nil)

有关的Rails 3.x的： Model.scoped

For Rails 3.x: Model.scoped

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