<p>为了实现从其json/yaml文件创建k8s资源的功能,我写了以下代码:</p>
<pre><code>def create_from_yaml(yaml_file):
"""
:param yaml_file:
:return:
"""
yaml_object = yaml.loads(common.load_file(yaml_file))
group, _, version = yaml_object["apiVersion"].partition("/")
if version == "":
version = group
group = "core"
group = "".join(group.split(".k8s.io,1"))
func_to_call = "{0}{1}Api".format(group.capitalize(), version.capitalize())
k8s_api = getattr(client, func_to_call)()
kind = yaml_object["kind"]
kind = re.sub('(.)([A-Z][a-z]+)', r'\1_\2', kind)
kind = re.sub('([a-z0-9])([A-Z])', r'\1_\2', kind).lower()
if "namespace" in yaml_object["metadata"]:
namespace = yaml_object["metadata"]["namespace"]
else:
namespace = "default"
try:
if hasattr(k8s_api, "create_namespaced_{0}".format(kind)):
resp = getattr(k8s_api, "create_namespaced_{0}".format(kind))(
body=yaml_object, namespace=namespace)
else:
resp = getattr(k8s_api, "create_{0}".format(kind))(
body=yaml_object)
except Exception as e:
raise e
print("{0} created. status='{1}'".format(kind, str(resp.status)))
return k8s_api
</code></pre>
<p>在上面的函数中,如果您提供了任何一个对象yaml/json文件,它会自动提取API类型和对象类型,并创建statefulset、deployment、service等对象</p>
<p>PS:上面的代码没有在一个文件中处理多个kubernetes资源,因此每个yaml文件应该只有一个对象。在</p>