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Docker links and required alias

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If no alias is passed one will now be created for you with the same name
as the target link container, as per the documentation
This commit is contained in:
Paul Myjavec 2014-06-10 22:30:27 +10:00
parent 8e3b512293
commit 074a6f3e3e
1 changed files with 17 additions and 2 deletions
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19
library/cloud/docker
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@ -380,7 +380,7 @@ class DockerManager:
self.links = None
if self.module.params.get('links'):
self.links = dict(map(lambda x: x.split(':'), self.module.params.get('links')))
self.links = self.get_links(self.module.params.get('links'))
self.env = None
if self.module.params.get('env'):
@ -391,6 +391,22 @@ class DockerManager:
self.client = docker.Client(base_url=docker_url.geturl())
def get_links(self, links):
"""
Parse the links passed, if a link is specified without an alias then just create the alias of the same name as the link
"""
processed_links = {}
for link in links:
parsed_link = link.split(':', 1)
if(len(parsed_link) == 2):
processed_links[parsed_link[0]] = parsed_link[1]
else:
processed_links[parsed_link[0]] = parsed_link[0]
return processed_links
def get_exposed_ports(self, expose_list):
"""
Parse the ports and protocols (TCP/UDP) to expose in the docker-py `create_container` call from the docker CLI-style syntax.
@ -452,7 +468,6 @@ class DockerManager:
return binds
def get_split_image_tag(self, image):
if '/' in image:
image = image.split('/')[1]