Hello.

I'm wondering if this data stucture is something you're getting back from 
another application or external source? If not, are you able to define the 
structure in anyway you desire? If not, why not?

Here is the code I managed to come up with, which has a much, much simpler 
data structure and does the same job.

---
- hosts: all
  sudo: no
  vars:
    applications:
      # - { cmd: 'bundle exec rake thing:1', path: 'application_1' }
      - { cmd: 'uptime', path: 'application_1' }
      - { cmd: 'uptime', path: 'application_2' }
      - { cmd: 'uptime', path: 'application_3' }
      - { cmd: 'uptime', path: 'application_4' }
  tasks:
    - name: Run dev tasks
      shell: >
        cd {{item.path}} && {{item.cmd}}
      with_items: applications



I hope this helps.

On Wednesday, 28 January 2015 02:11:01 UTC, Evan Tahler wrote:
>
> Hi All!  I cannot sort out how to itterate though this loop properly:
>
> I have 2 data dictionaries: 
>
> applications:
>   - application_1
>   - application_2
>
> and 
>
> big_var_collection:
>   application_1:
>     deployment_tasks:
>       - { cmd: 'bundle exec rake thing:1', metadata: 'yay' }
>       - { cmd: 'bundle exec rake thing:2', metadata: 'boo' }
>   application_2:
>     deployment_tasks:
>       - { cmd: 'bundle exec rake thing:3', metadata: 'yay' }
>       - { cmd: 'bundle exec rake thing:4', metadata: 'boo' }
>
> In this example, I want to run 4 commands: cd into the directory and run 
> the `cmd`, IE:
>
> cd ~/www/projects/application_1 && bundle exec rake thing:1
> cd ~/www/projects/application_1 && bundle exec rake thing:2
> cd ~/www/projects/application_2 && bundle exec rake thing:3
> cd ~/www/projects/application_2 && bundle exec rake thing:4
>
> I would have assumed that the following loop would work, but I always end 
> up with an access error of some sort
>
> - name: run deployment tasks
>   shell: > 
>     cd ~/www/projects/{{ item.0 }} && {{ item.1.cmd }}
>   with_nested: 
>   - applications
>   - big_var_collection[item.0].deployment_tasks
>
> Help?
>

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