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Remove content view version from environments if empty #1280

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@ehelms ehelms commented Aug 26, 2021

Probably needs some testing added, did not want to forget about this change I encountered doing some cleanup tasks.

current_environment_ids = {environment['id'] for environment in content_view_version['environments']}

for environment_id in current_environment_ids:
response = module.resource_action('content_views', 'remove_from_environment', params={'id': content_view['id'], 'environment_id': environment_id})
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@ehelms ehelms Aug 26, 2021

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On newer releases of Katello, I think this can be done in one call via:

response = module.resource_action('content_views_versions', 'remove_from_environments', params={'id': content_view['id'], 'environment_ids': current_environment_ids})

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Newer is 4.0+? If there is a possible perfomance gain, I'd be in favor of an if/else?

BTW, the linter says this line is too damn long (170>160 chars) ;)

force=module.foreman_params['force_promote'],
force_yum_metadata_regeneration=module.foreman_params['force_yum_metadata_regeneration'],
)
if module.foreman_params['lifecycle_environments'] == [] and content_view_version['environments']:
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Should this happen only if the user says [] or always when the list doesn't match?

The later has some complications as we never list Library and I think the current user expectation is "promote to these envs, don't touch the others"?

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