Rails 3.如何获得两个数组之间的区别？

Question

比方说，我有这个数组与发货ID。发票与货物ID的

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

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

• 发票

  s = Shipment.find(:all, :select => "id") 
  
  [#<Shipment id: 1>, #<Shipment id: 2>, #<Shipment id: 3>, #<Shipment id: 4>, #<Shipment id: 5>] 
  

  阵列属于装运。

• 装运有一个发票。
• 所以发票表有一列shipment_id。

要创建一个发票，我点击新发票，然后有一个选择菜单与发货，所以我可以选择“我要为哪个发货创建发票”。所以我只想显示尚未创建发票的货件清单。

所以我需要一个没有发票的货件数组呢。在上面的例子中，答案是1，4，5

Answer 1

首先，你将得到的是出现在发票shipping_id的列表：

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

然后从原始数组“拒绝”他们：

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

注：记住，拒绝返回一个新数组，使用拒绝！如果你想改变原有的阵列

Answer 2

使用替代标志

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

Answer 3

这里从pgquardiario以前的答案只包括一个方向差异的。如果你想要两个数组的区别（就像它们都有一个独特的项目），那么就试试下面的内容。

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 4

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

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

Answer 5

这应做到在一个ActiveRecord的查询

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

，并将其输出的SQL

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

---

在的Rails 4+你可以做跟着

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

，并将其输出的SQL

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

，而是select(:id)我建议ids方法。

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