How to render a view/partial in a helper w/o the layout?

Question

Hello in a helper I'm doing the following:

render(:template =>"feeds/_feed_item.html.erb", :locals => { :feed_item => feed_item }).to_s

Problem is this is rendering the layout which I don't want. How can I render just the file/template feed_item ?

Thanks

score 2 · Accepted Answer · edited Feb 07 '13 at 17:26

2

You can add the option :layout => false to render without the layout.

Example:

render(:template =>"feeds/_feed_item.html.erb", :layout => nil , :locals => { :feed_item => feed_item }).to_s

edited Feb 07 '13 at 17:26

fresskoma

answered Jun 12 '11 at 00:17

Devin M

Can you render this in a model somehow? – AnApprentice Jun 12 '11 at 02:51
1

Try using a partial for this. http://guides.rubyonrails.org/layouts_and_rendering.html#using-partials – Devin M Jun 12 '11 at 02:59
It must be `layout => nil`, at least in Rails 3.1! Edited accordingly. – fresskoma Feb 07 '13 at 17:25

score 1 · Answer 2 · answered Jun 12 '11 at 02:32

1

If you use the :partial key, the template shouldn't be rendered. Try this:

render :partial => 'feeds/feed_item', :locals => { :feed_item => feed_item }).to_s

Depending on your setup, you may also be able to shorten this to simply:

render feed_item

answered Jun 12 '11 at 02:32

dpb

Can you render this in a model somehow? – AnApprentice Jun 12 '11 at 02:52
I've never done it, but this looks promising. http://tumblr.seebq.com/post/39566909/rails-code-render-a-view-from-a-model-require – dpb Jun 12 '11 at 02:59
1

There's a trailing parenthesis in this answer, can you remove it? – Joshua Plicque Jan 12 '15 at 20:12

2 Answers2