8

我正在关注 Bate 在图表上的 tut，但我遇到了一个错误。

这是我得到的错误

PG::Error: ERROR:  column "orders.created_at" must appear in the GROUP BY clause or be used in an aggregate function
LINE 1: SELECT created_at, sum(amount) as total_amount FROM "orders"...
               ^
: SELECT created_at, sum(amount) as total_amount FROM "orders"  WHERE ("orders"."created_at" BETW

这是我在 Orders 模型中使用的代码

def self.total_grouped_by_day(start)
    orders = where(created_at: start.beginning_of_day..Time.zone.now)#.all
    orders = orders.group("date(created_at)")
    orders = orders.select("date(created_at), sum(amount) as total_amount")
    orders.group_by { |order| order.created_at.to_date }
  end

在我的助手中

def orders_chart_data
        orders_by_day = Order.total_grouped_by_day(3.weeks.ago)
        (3.weeks.ago.to_date..Date.today).map do |date|
            {
                created_at: date,
                price: orders_by_day[date].first.try(:total_amount) || 0
            }
        end
    end

我正在使用 Postgres（我很确定这是我收到此错误的主要原因）和 Rails 3.xx 为什么会发生此错误以及如何解决？

先感谢您

在我的控制台中：

Order.where(created_at: 3.weeks.ago.beginning_of_day..Time.zone.now).group("date(created_at)").select("date(created_at), sum(amount) as total_amount")

这就是我得到的

[#<Order >, #<Order >, #<Order >]

奇怪的..

在我的控制台中运行它：Order.where(created_at: 3.weeks.ago.beginning_of_day..Time.zone.now).group("date(created_at)").select("date(created_at), sum(amount) as total_amount").first.total_amount

给了我这个：

Creating scope :page. Overwriting existing method Order.page.
  Order Load (18.1ms)  SELECT date(created_at), sum(amount) as total_amount FROM "orders" WHERE ("orders"."created_at" BETWEEN '2013-07-05 00:00:00.000000' AND '2013-07-26 23:50:38.876609') GROUP BY date(created_at) LIMIT 1
 => "10.0"
4

2 回答 2

6

您正在分组date(created_at)但未选择该列。更改此行：

orders = orders.select("created_at, sum(amount) as total_amount")

对此：

orders = orders.select("date(created_at), sum(amount) as total_amount")

这也将导致对下一行的 group_by 进行更改。

从我的一个项目中，使用略有不同的属性，但做的事情和你一样：

1.9.3p327 > User.group('date(created_at)').select('date(created_at), sum(id) as total_amount').first.attributes
  User Load (1.2ms)  SELECT date(created_at), sum(id) as total_amount FROM "users" GROUP BY date(created_at) LIMIT 1
 => {"date"=>"2011-09-27", "total_amount"=>"657"}
于 2013-07-26T22:54:08.907 回答
1

迄今为止的演员阵容将是最简单和最快的。这里是原始 SQL，因为我不是 Ruby 语法专家（并且一般不喜欢 ORM 屠杀 SQL）。

SELECT created_at::date, sum(amount) AS total_amount
FROM   orders
WHERE  created_at ...
GROUP  BY created_at::date
ORDER  BY created_at::date
于 2013-07-26T23:07:43.140 回答