Rails/Arel:选择所有记录作为 ActiveRecord::Relation [英] Rails/Arel: Selecting all records as an ActiveRecord::Relation

问题描述

在 Rails 中使用 Arel - 我正在寻找一种创建 ActiveRecord::Relation 的方法，它可以有效地生成 SELECT * FROM table，我仍然可以操纵它进一步.

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，我有一个分为多个类别的模型，我通过以下方式返回这些类别的计数:

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
}

这很好用，并且具有向 MySQL 发送 COUNT 查询的优点，而不是实际选择记录本身.

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

但是，我现在需要在计数中包含存档记录，但是 relation = Model.all 会生成一个 Array，我正在寻找一个 ActiveRecord::Relation.

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.

我能想到的唯一方法是 model.where(model.arel_table[:id].not_eq(nil))，它有效，但似乎有点荒谬.

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.

任何人都可以对此有所了解吗?

Can anyone shed any light on this?

推荐答案

对于 Rails 4.1 及更高版本:Model.all 返回一个关系(以前没有)

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

对于 Rails 4.0:Model.where(nil)

For Rails 4.0: Model.where(nil)

对于 Rails 3.x:Model.scoped

For Rails 3.x: Model.scoped

这篇关于Rails/Arel:选择所有记录作为 ActiveRecord::Relation的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！