I've got the following template file creation in my cookbook:
template "my_file" do
path "my_path"
source "my_file.erb"
owner "root"
group "root"
mode "0644"
variables(#template_variables)
notifies :restart, resources(service: "my_service")
end
and the following assertions in my ChefSpec tests:
chef_run.should create_file "my_file"
chef_run.file("my_file").should be_owned_by('root', 'root')
Which results in the following failure:
No file resource named 'my_file' with action :create found.
This is due to the fact that I am not using afile resource but a template resource. Question: How can I test for file creation off a template resource using ChefSpec?
There are two ways to solve your problem.
First, you can use the create_template matcher. This will match only "template" resources in the run context:
expect(chef_run).to create_template('my_file')
This matcher is also chainable, so you can assert attributes:
expect(chef_run).to create_template('my_file')
.with_path('my_path')
.with_owner('root')
However, this matcher won't actually render the template. So you can't check if you've setup file-specificity correctly.
There's also a top-level matcher for any kind of "file" (file, cookbook_file, and template) that actually renders the contents in memory:
expect(chef_run).to render_file('my_file').with_content(/^match me$/)
You can find more information about render_file in the README.
According to the docs (https://github.com/acrmp/chefspec) you should be able to use:
expect(chef_run).to create_file 'my_file'
I think something changed very recently (possibly the version of chefspec on rubygems), however, because tests I had passing earlier today (using the same syntax you are using) are now suddenly failing.
Related
I have a Chef cookbook with many recipes that have the same code, beside other particular things.
template 'stack_file' do
local true
source File.join(base_dir, 'stack_templates/admin.yml.erb')
path File.join(base_dir, 'stacks/admin.yml')
variables(context)
end
template 'settings_file' do
sensitive true
local true
source File.join(base_dir, 'config_templates/settings_admin.yml.erb')
path File.join(base_dir, 'configs/settings_admin.yml')
variables(context)
end
Is it possible to somehow put this code in a method that I would call with my source_file, destination_file and variables?
I guess you can write a module and include it in your recipes as you do in plain Ruby.
module StackFile
...the code you want to share...
end
Then you can use:
inlclude StackFile
or
Chef::Recipe.send(:include, StackFile)
or when using *_if conditions
Chef::Resource.send(:include, StackFile)
Do this:
create new cookbook.
don't create recipes in it, but instead create a resource
you can define any parameters (inputs like source file, dest file etc like you mentioned)
add your templates to the new cookbook.
in your other cookbooks, create dependency to the previously created cookbook. This will enable you to call the resource you created there (remember that when you call resource from another cookbook and it is creating templates, it will try to take the template file from the current cookbook and not the one where the resource is defined. That is why you need to specify cookbook name when creating template (in the shared cookbook's resource) - see cookbook attribute of https://docs.chef.io/resource_template.html)
Repeat for any number of cookbooks.
I'm deploying a package that requires a template be created in a specified directory every time a directory is unzipped.
A remote_file notifies my unzip action, that unzip action notifies the template resource, which in turn notifies other resources. This chain of notifications works as expected.
Below is my template resource:
template 'C:\\Program Files\\MyProgram\\program.yml' do
source "my_program-#{node['program']['version']}.yml.erb"
action :nothing
notifies :run, 'powershell_script[install-program]', :immediately
end
My question: Is there a way to have the template resource execute if I make a change to the source template? Right now it only executes the template resource if notified by my unzip action (due to my action :nothing).
However, it would be great to have a way for it to tell if the template itself has changed. Perhaps some kind of not_if or only_if statement?
sounds to me that you avoid all the notification chaining if you will have your resources defined in the same recipe.
back to your questions, it sounds that setting action :create, which is the default action, will do the trick. from the template resource documentation
action :create
Create a file. If a file already exists (but does not match), update that file to match.
I am trying to create an LWRP that will call the resource that is defined within itself. My cookbook's structure is as follows:
In the machine cookbook's provider, I have a code snippet as follows:
require 'chef/provisioning' # driver for creating machines
require '::File'
def get_environment_json
##environment_template = JSON.parse(File::read(new_resource.template_path + "environment.json"))
return ##environment_template
end
The code is only trying to read a json file and I am using File::read for it.
I keep getting an error as follows:
LoadError
cannot load such file -- ::File
Does anyone know how I can use File::read inside my LWRP's provider?
OK, so the prior two answers are both half right. You have two problems.
First, you can't require ::File as it's already part of Ruby. This is the cause of your error.
Second, if you call File.read you will grab Chef's File not ruby's. You need to do a ::File.read to use Ruby's File class.
require '::File'
Is incorrect and is causing the LoadError. Delete this line. You don't need it. File is part of the Ruby core and doesn't need to be required.
To further explain, the string argument to require represents the file name of the library you want to load. So, it should look like require "file", or require "rack/utils".
It happens becuase Chef already has a file resource. We have to use the Ruby File class in a recipe.We use ::File to use the Ruby File class to fix this issue. For example:
execute 'apt-get-update' do
command 'apt-get update'
ignore_failure true
only_if { apt_installed? }
not_if { ::File.exist?('/var/lib/apt/periodic/update-success-stamp') }
end
Source: https://docs.chef.io/ruby.html#ruby-class
I would like to read a file from a checkout git repository to parse a config file and use this data to perform few resources commands.
git "/var/repository" do
action :sync
end
config = JSON.parse(File.read("/var/repository/config.json" ))
config.each do |job, flags|
#do some resources stuff here
end
This will not work because the file doesn't exist at compile time:
================================================================================
Recipe Compile Error in /var/chef/cache/cookbooks/...
================================================================================
Errno::ENOENT
No such file or directory - /var/repository/config.json
I where trying to load the file in ruby_block and perform the Chef resource actions there, but this didn't worked. Also setting the parsed config to a variable and use it outside of the ruby_block didn't work.
ruby_block "load config" do
block do
config = JSON.parse(File.read("/var/repository/config.json"))
#node["config"] = config doesn't work - node["config"] will not be set
config.each do |job, flags|
#do some stuff - will not work because Chef context is missing
end
end
end
Any idea how I could read the file at runtime and used the parsed values in my recipe?
You may also find it helpful to use lazy evaluation in scenarios like this.
In some cases, the value for an attribute cannot be known until the execution phase of a chef-client run. In this situation, using lazy evaluation of attribute values can be helpful. Instead of an attribute being assigned a value, it may instead be assigned a code block.
I am trying to follow this tutorial:
http://net.tutsplus.com/tutorials/ruby/singing-with-sinatra/
Got stuck in "We’ll also make use of a “view file”, which allows us to split the markup for a view into a separate file. "
I have my basics.rb file running fine.
And My files are stored as follows:
Desktop/RubyForm/basics.rb
Desktop/RubyForm/view/form.erb
However, now when i go to http://localhost:9393/form , I am greeted with:
Errno::EIO at /form
Input/output error - <STDERR> file: lint.rb location: write line: 398
sinatra.error
Errno::ENOENT: No such file or directory -
/Users/HelenasMac/Desktop/views/form.erb
UPDATE! : Got the form to work right after running ruby basics.rb and going to http://localhost:4567/form .
However, after I run "shotgun basics.rb" , I have to go to
http://localhost:9393/form, and that's when the form doesn't show up.
What am I doing wrong? Disclaimer: mega beginner to ruby and using the terminal.
Thanks in advance!
If you cannot get shotgun to work then the new recommended way to reload Sinatra seems to be rerun.
To use it:
> gem install rerun
> cd /Users/HelenasMac/Desktop/RubyForm
> rerun ruby basics.rb
Explicity Set a Views Directory
Unless you're using inline template for your views with enable :inline_templates, you may need to explicitly define a template directory if the default values aren't working for you. The docs describe how to set your views directory as follows:
:views - view template directory
A string specifying the directory where view templates are located. By default, this is assumed to be a directory named “views” within the application’s root directory (see the :root setting). The best way to specify an alternative directory name within the root of the application is to use a deferred value that references the :root setting:
set :views, Proc.new { File.join(root, "templates") }
You may also need to explicitly set :root, and make sure that both :root and :views make sense from your current working directory.