Ruby on Rails generate model

Question

So, I want to make my models render to json object so it can be used for a Android Device. I have all the models setup, but because I used the sccafold generate model call, I don't have any methods in the controller of the models... How do I generate the actual code, so that I can render the model in json?

I am using

$ rails --version
Rails 3.2.13

The answer can easily be found with a quick google search.

ericeason
– ericeason

2013-04-25 01:33:01 +00:00
Commented Apr 25, 2013 at 1:33 — ericeason
– ericeason, Commented Apr 25, 2013 at 1:33

Community · Accepted Answer · 2017-05-23 11:57:19Z

1

From the Ruby on Rails guide:

render :json => @product

So, inside the generated controller you can do something like this:

def show
    @user = User.find( params[:id] )
    respond_to do |format|
        format.json{
            render :json => @user.to_json
        }
    end
end

Tip: You don’t need to call to_json on the object that you want to render. If you use the :json option, render will automatically call to_json for you.

You can find some examples on questions that have been answered previously:

edited May 23, 2017 at 11:57

CommunityBot

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answered Apr 25, 2013 at 1:34

Diego Cardozo

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2 Comments

Sriram Venkatesh Over a year ago

yeah but where do I put it if the code generated from scafollding is: class PostsController < InheritedResources::Base before_filter :authenticate_user! end

Diego Cardozo Over a year ago

I edited the answer and added some code to give you a better idea of where this code is supposed to go.

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