Question

我有一系列哈希，我试图将其植入数据库。

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

我只为department_id和follower_id获取空值。 user_id正在运作。

我尝试使用"#{n}"和"#{k}"来获取设置为部门和关注者ID的键值。我还尝试仅使用.each do |a|并设置department_id: a['department_id'], follower_id a['follower_id']

来迭代数组

如此处所示：iterate through array of hashes in ruby和此处：How do I iterate over an array of hashes and return the values in a single string?

但我仍然只能获得空值。如何将我的值输入数据库？

Answer 1

shoe_array是一个散列数组，因此您应该迭代每个散列，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create来自一系列哈希的记录：

以下应该可以使用（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|的示例可以是散列或数组数组。如果你想沿着那条路走下去，你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始终在follower_id之前出现）

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]]

然后您可以使用旧代码或重构

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

请注意，虽然您正在迭代数组两次，但与第一次相比效率会降低。

<强>更新

另一种选择是按原样使用数组元素。

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

迭代哈希数组以在rails中创建数据

3 个答案: