迭代一系列哈希，以在导轨中创建数据

Question

我有一系列我试图在数据库中播种的哈希。

shoe_array = [{:department_id=>8, :follower_id=>31}, {:department_id=>9, :follower_id=>41}, {:department_id=>4, :follower_id=>49}, {:department_id=>2, :follower_id=>58}, {:department_id=>5, :follower_id=>36}, {:department_id=>9, :follower_id=>63}, {:department_id=>2, :follower_id=>52}, {:department_id=>23, :follower_id=>26}, {:department_id=>5, :follower_id=>52}, {:department_id=>6, :follower_id=>30}]

shoe_array.each do |n, k|
  department_id = n,
  follower_id = k,
  user_id = 1
  Relationship.create!(department_id: department_id,
                       follower_id: follower_id,
                       user_id: user_id)
end

我仅获得department_id和follower_id的零值。user_id正在工作。

我尝试使用"#{n}"和"#{k}"，以将键值设置为部门和追随者ID。我还尝试仅使用.each do |a|和设置department_id: a['department_id'], follower_id a['follower_id']

在数组上迭代迭代

如下所示：在Ruby中迭代一系列的哈希和这里：如何迭代一系列哈希并返回一个字符串中的值？

，但我仍然只得到无效的值。我如何将我的值纳入数据库？

Answer 1

shoe_array是一系列哈希，因此您应该在每个哈希上迭代，并访问每个键值对：

shoe_array.each do |hash|
  department_id = hash[:department_id]
  follower_id   = hash[:follower_id]
  user_id       = 1
  Relationship.create!(
    department_id: department_id,
    follower_id:   follower_id,
    user_id:       user_id
  )
end

Answer 2

根据文档，您可以从一系列哈希创建记录：

以下应该有效（您可以使用create!以及create）

shoe_array = [{:department_id=>8, :follower_id=>31}, {:department_id=>9, :follower_id=>41}, {:department_id=>4, :follower_id=>49}, {:department_id=>2, :follower_id=>58}, {:department_id=>5, :follower_id=>36}, {:department_id=>9, :follower_id=>63}, {:department_id=>2, :follower_id=>52}, {:department_id=>23, :follower_id=>26}, {:department_id=>5, :follower_id=>52}, {:department_id=>6, :follower_id=>30}]
Relationship.create!(shoe_array.map{|arr| arr.merge!({user_id: 1})})

Answer 3

将迭代更改为

shoe_array.each do |shoe|
  department_id = shoe[:department_id]
  follower_id = shoe[:follower_id]

可以使用|n, k|的示例是哈希或数组数组。如果您想沿着该路线走下去，可以在数组中的每个哈希上调用values（假设哈希是一致的，这意味着department_id总是在follower_id之前先出现在第一个）

ids = shoe_array.map(&:values) # [[8, 31], [9, 41], [4, 49], [2, 58], [5, 36], [9, 63], [2, 52], [23, 26], [5, 52], [6, 30]]

然后，您只能将旧代码或重构来

ids.each do |department_id, follower_id|
  Relationship.create!(
    department_id: department_id,
    follower_id: follower_id,
    user_id: 1
  )
end

请注意，尽管您在数组上迭代两次，并且与第一个相比，效率也不太效率。

update

另一个选项是使用数组元素。

shoe_array.each do |attributes|
  relationship = Relationship.new(attributes)
  relationship.user_id = 1
  relationship.save!
end