问题
I'm not sure if I'm doing the best approach here, but I have a block of data that I want to show after a search is done and to not be there at all before. First of all, there is nothing to show, and second the model it references is nil so it throws an exception.
I placed this block in a partial template and added it the appropriate spot in my layout. Is there a way to cleanly render the partial conditionally? Is there a better way to approach this problem?
回答1:
Ruby allows you to do nice things like this:
<%= render :partial => "foo/bar" if @conditions %>
To make this a bit easier to read and understand, it can be written as:
<%= render(:partial => "foo/bar") if @conditions %>
render is a function, and you pass it a hash that tells it which partial to render. Ruby allows you to put things on one line (which often makes them more readable and concise, especially in views), so the if @conditions section is just a regular if statement. It can also be done like:
<% if @conditions %>
<%= render :partial => "foo/bar" %>
<% end %>
Edit:
Ruby also allows you to use the unless keyword in place of if. This makes code even more readable, and stops you from having to do negative comparisons.
<%= render :partial => "foo/bar" if !@conditions %>
#becomes
<%= render :partial => "foo/bar" unless @conditions %>
回答2:
One easy way is to use a helper method. Helpers tend to be a bit cleaner than putting logic directly in the view.
So, your view might be something like :
<%= render_stuff_conditionally %>
and your helper would have a method to control this:
def render_stuff_conditionally
if @contional_check
render :partial => 'stuff'
end
end
where obviously things are named more appropriately
回答3:
Assuming I am following you right, you do this at the view level.
<% if !@my_search_data.nil? %>
<% render :partial => 'foo/bar' %>
<% end %>
Hope that helps. If not, maybe post an example of your code.
来源：https://stackoverflow.com/questions/2376482/ruby-on-rails-conditionally-display-a-partial