ruby或rails中是否有序数到基数函数？

Question

我正在尝试找到一种更好的方式来表达我的黄瓜，所以我正在寻找一个序数到基数函数，将其转换

When I fill up the first passenger field
Then I should see the passenger list update with the first passenger details
When I follow "Add Another Passenger"
Then I should see a second passenger field
When I fill up the second passenger field
Then I should see the passenger list update with the second passenger details

为更动态的东西（而不是为每行创建单独的步骤）

这是我的网络步骤的示例

When /^I fill up the first passenger field$/ do
  fill_in("booking_passengers_attributes_0_first_name", :with => "Blah")
  fill_in("booking_passengers_attributes_0_last_name", :with => "blah")
  select("5' to 6'", :from => "booking_passengers_attributes_0_height")
  select("100 to 150lbs", :from => "booking_passengers_attributes_0_weight")
end

When /^I fill up the second passenger field$/ do
  fill_in("booking_passengers_attributes_1_first_name", :with => "Wee")
  fill_in("booking_passengers_attributes_1_last_name", :with => "Sir")
  select("5' to 6'", :from => "booking_passengers_attributes_1_height")
  select("150 to 200lbs", :from => "booking_passengers_attributes_1_weight")
end

请参阅0 和 1？我希望将“第一个”转换为基数，这样我就可以替换。您也可以建议一种更好的方式来声明 cukes :)

更新的答案 我正处于重构过程中，但基本上我使用了 1st 而不是first，并使用了 to_i 。

When /^I fill up the "([^"]*)" passenger field$/ do |id|
  input_id = id.to_i - 1
  fill_in("booking_passengers_attributes_#{input_id}_first_name", :with => id)
  fill_in("booking_passengers_attributes_#{input_id}_last_name", :with => "Passenger")
  select("5' to 6'", :from => "booking_passengers_attributes_#{input_id}_height")
  select("100 to 150lbs", :from => "booking_passengers_attributes_#{input_id}_weight")  
end

Answer 1

我并不完全理解，确切地说你想要做什么，但是你可以在积极支持的情况下做这样的事情：

 1.ordinalize    # => "1st"
  2.ordinalize    # => "2nd"
  1002.ordinalize # => "1002nd"

并且有一个动作视图助手 number_in_words 来获取“第一”，“第二”等

我对 cukes 不太了解对不起，

Answer 2

使用简短、易于解析的序数词：

When /^I fill up the (\d+)(?:st|nd|rd|th) passenger field$/ do |n|
  # etc...
end

Answer 3

此函数内置于 Chronic 中：

irb(main):001:0> require 'chronic'
=> true
irb(main):002:0> Chronic::Numerizer.numerize("eighty-fifth").to_i
=> 85