问题
How would you make an erb template that has human readable json?
The following code works, but it makes a flat json file
default.rb
default['foo']['bar'] = { :herp => 'true', :derp => 42 }
recipe.rb
template "foo.json" do
source 'foo.json.erb'
variables :settings => node['foo']['bar'].to_json
action :create
end
foo.json.erb
<%= @settings %>
Similar SO questions
Chef and ruby templates - how to loop though key value pairs?
How can I "pretty" format my JSON output in Ruby on Rails?
回答1:
As pointed out by this SO Answer .erb templates are great for HTML, and XML, but is not good for json.
Luckily, CHEF uses its own json library which has support for this using .to_json_pretty
@coderanger in IRC, pointed out that you can use this library right inside the recipe. This article shows more extensively how to use chef helpers in recipes.
default.rb
# if ['foo']['bar'] is null, to_json_pretty() will error
default['foo']['bar'] = {}
recipe/foo.rb
pretty_settings = Chef::JSONCompat.to_json_pretty(node['foo']['bar'])
template "foo.json" do
source 'foo.json.erb'
variables :settings => pretty_settings
action :create
end
Or more concise as pointed out by YMMV
default.rb
# if ['foo']['bar'] is null, to_json_pretty() will error
default['foo']['bar'] = {}
recipe/foo.rb
template "foo.json" do
source 'foo.json.erb'
variables :settings => node['foo']['bar']
action :create
end
templates/foo.json.erb
<%= Chef::JSONCompat.to_json_pretty(@settings) %>
回答2:
Something like this would also work:
file "/var/my-file.json" do
content Chef::JSONCompat.to_json_pretty(node['foo']['bar'].to_hash)
end
来源:https://stackoverflow.com/questions/31083484/how-do-you-create-pretty-json-in-chef-ruby