Simple optimization of creating/updating workflows
In case we load definition of one workflow, there is no need to convert it to yaml and back. Change-Id: Ia7fb914843e56650c8f6ff5e8c16d1b57001c0fc Signed-off-by: Oleg Ovcharuk <vgvoleg@gmail.com>
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@ -91,15 +91,20 @@ def create_workflows(definition, scope='private', is_system=False,
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def _append_all_workflows(definition, is_system, scope, namespace,
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wf_list_spec, db_wfs):
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wfs = wf_list_spec.get_workflows()
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wfs_yaml = yaml.load(definition)
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wfs_yaml = yaml.load(definition) if len(wfs) != 1 else None
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for wf_spec in wfs:
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if len(wfs) != 1:
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definition = _cut_wf_definition_from_all(
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wfs_yaml,
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wf_spec.get_name()
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)
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db_wfs.append(
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_create_workflow(
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wf_spec,
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_cut_wf_definition_from_all(
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wfs_yaml,
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wf_spec.get_name()
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),
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definition,
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scope,
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namespace,
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is_system
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@ -125,19 +130,25 @@ def update_workflows(definition, scope='private', identifier=None,
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db_wfs = []
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wfs_yaml = yaml.load(definition) if len(wfs) != 1 else None
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with db_api.transaction():
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wfs_yaml = yaml.load(definition)
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for wf_spec in wfs:
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db_wfs.append(_update_workflow(
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wf_spec,
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_cut_wf_definition_from_all(
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if len(wfs) != 1:
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definition = _cut_wf_definition_from_all(
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wfs_yaml,
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wf_spec.get_name()
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),
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scope,
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namespace=namespace,
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identifier=identifier
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))
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)
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db_wfs.append(
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_update_workflow(
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wf_spec,
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definition,
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scope,
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namespace=namespace,
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identifier=identifier
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)
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)
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return db_wfs
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