Ruby on Rails - Simple calculations shown in the view and stored in database

Question

First of all sorry for this simple question, after spending a lot of time I have given up finding it without any help :(.

Simplified I have the following situation: one table rates with columns rate and converted generated with the scaffold statement within ruby. The column rate will be shown in the view, but the converted column not which will only be used in the background of the application.

Now it's my goal to ask the user for a rate given in the field rate so the application can calculate the converted and store this value in the database. It's a simple calculation: converted = 1 / rate.

Please advice :).

AJcodez · Accepted Answer · 2012-07-30 08:30:03Z

3

Well there are a ton of ways to do that, including but not limited to putting the logic in the model (probably cleanest), in the controller and in the view itself with a hidden field and javascript.

I would do it at the model level with a callback. For example, add this to your rates.rb model:

before_save :calculate_converted_rate

private

def calculate_converted_rate
  self.converted = 1.0 / self.rate
end

Which as the code might suggest sets the converted field to the inverse of the rate field each time before a Rates instance is saved.

More information on ActiveRecord callbacks: http://api.rubyonrails.org/classes/ActiveRecord/Callbacks.html

answered Jul 30, 2012 at 8:30

AJcodez

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Comments

Edd Morgan · Accepted Answer · 2012-07-30 08:28:47Z

2

Define your own mutator method in your model. This will allow you to take the input from your user, do what you want with it, then store it in the database. To illustrate:

class Rate < ActiveRecord::Base

    def rate=(value)
        write_attribute :rate, value
        write_attribute :converted, 1.0 / value
    end

end

This will override Rails' default rate= method, which is what gets called when your model is passed your rate data from (assumedly) your form. Construct your model's form normally, with a <%= f.text_field :rate %> or whatever you need to do.

Note the 1.0 / value, not 1 / value - this will ensure the result of that expression is a float, and that you won't lose any precision. If you divide an integer by an integer in Ruby, you'll always get an integer.

answered Jul 30, 2012 at 8:28

Edd Morgan

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2 Comments

jmk Over a year ago

Thanks a lot, I will try this solution when the application needs more calculations.

ecoologic Over a year ago

+1 I like this solutions 10 times more, the value is always updated, callbacks will leave you with wrong data in the instance

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