Rails 3.如何获得两个数组之间的差异？

Question

假设我有这个带有出货ID的数组.

s = Shipment.find(:all, :select => "id")

[#<Shipment id: 1>, #<Shipment id: 2>, #<Shipment id: 3>, #<Shipment id: 4>, #<Shipment id: 5>]

带有货件ID的发票数组

i = Invoice.find(:all, :select => "id, shipment_id")

[#<Invoice id: 98, shipment_id: 2>, #<Invoice id: 99, shipment_id: 3>]

• 发票属于Shipment.
• 装运有一个发票.
• 因此,发票表有一列shipment_id.

要创建发票,我点击新发票,然后有一个包含发货的选择菜单,所以我可以选择"我为哪个发货创建发票".所以我只想显示尚未为其创建发票的货件清单.

所以我需要一组没有发票的货件.在上面的例子中,答案是1,4,5.

Answer 1

a = [2, 4, 6, 8]
b = [1, 2, 3, 4]

a - b | b - a # => [6, 8, 1, 3]

Answer 2

首先,您将获得发票中显示的shipping_id列表:

ids = i.map{|x| x.shipment_id}

然后从原始数组中"拒绝"它们:

s.reject{|x| ids.include? x.id}

注意:记住拒绝返回一个新数组,使用拒绝!如果要更改原始数组

Answer 3

使用替代标志

irb(main):001:0> [1, 2, 3, 2, 6, 7] - [2, 1]
=> [3, 6, 7]

Answer 4

Ruby 2.6引入了Array.difference：

[1, 1, 2, 2, 3, 3, 4, 5 ].difference([1, 2, 4]) #=> [ 3, 3, 5 ]

所以在这里给出的情况下：

Shipment.pluck(:id).difference(Invoice.pluck(:shipment_id))

似乎是解决问题的一种很好的解决方案。我一直是的热心追随者a - b | b - a，尽管有时想起来可能很棘手。

这当然可以解决这个问题。

Answer 5

纯红宝石溶液是

(a + b) - (a & b)

([1,2,3,4] + [1,3]) - ([1,2,3,4] & [1,3])
=> [2,4]

哪里a + b会产生两个阵列之间的联合
与a & b回报交集
而union - intersection会返回差异

Answer 6

此前pgquardiario的答案仅包括单向差异.如果你想要两个数组的差异(因为它们都有一个唯一的项目),那么尝试类似下面的内容.

def diff(x,y)
  o = x
  x = x.reject{|a| if y.include?(a); a end }
  y = y.reject{|a| if o.include?(a); a end }
  x | y
end

Answer 7

这应该在一个ActiveRecord查询中执行

Shipment.where(["id NOT IN (?)", Invoice.select(:shipment_id)]).select(:id)

它输出SQL

SELECT "shipments"."id" FROM "shipments"  WHERE (id NOT IN (SELECT "invoices"."shipment_id" FROM "invoices"))

在Rails 4+中,您可以执行以下操作

Shipment.where.not(id: Invoice.select(:shipment_id).distinct).select(:id)

它输出SQL

SELECT "shipments"."id" FROM "shipments"  WHERE ("shipments"."id" NOT IN (SELECT DISTINCT "invoices"."shipment_id" FROM "invoices"))

而不是select(:id)我推荐的ids方法.

Shipment.where.not(id: Invoice.select(:shipment_id).distinct).ids