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53 lines
2.1 KiB
Python
53 lines
2.1 KiB
Python
"""
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* *******************************************************
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* Copyright (c) VMware, Inc. 2016. All Rights Reserved.
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* SPDX-License-Identifier: MIT
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* *******************************************************
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*
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* DISCLAIMER. THIS PROGRAM IS PROVIDED TO YOU "AS IS" WITHOUT
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* WARRANTIES OR CONDITIONS OF ANY KIND, WHETHER ORAL OR WRITTEN,
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* EXPRESS OR IMPLIED. THE AUTHOR SPECIFICALLY DISCLAIMS ANY IMPLIED
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* WARRANTIES OR CONDITIONS OF MERCHANTABILITY, SATISFACTORY QUALITY,
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* NON-INFRINGEMENT AND FITNESS FOR A PARTICULAR PURPOSE.
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"""
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__author__ = 'VMware, Inc.'
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__copyright__ = 'Copyright 2016 VMware, Inc. All rights reserved.'
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__vcenter_version__ = '6.5+'
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from com.vmware.vcenter_client import VM
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from samples.vsphere.vcenter.helper import datastore_helper
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from samples.vsphere.vcenter.helper import folder_helper
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from samples.vsphere.vcenter.helper import resource_pool_helper
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def get_placement_spec_for_resource_pool(stub_config,
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datacenter_name,
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vm_folder_name,
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datastore_name):
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"""
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Returns a VM placement spec for a resourcepool. Ensures that the
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vm folder and datastore are all in the same datacenter which is specified.
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"""
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resource_pool = resource_pool_helper.get_resource_pool(stub_config,
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datacenter_name)
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folder = folder_helper.get_folder(stub_config,
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datacenter_name,
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vm_folder_name)
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datastore = datastore_helper.get_datastore(stub_config,
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datacenter_name,
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datastore_name)
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# Create the vm placement spec with the datastore, resource pool and vm
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# folder
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placement_spec = VM.PlacementSpec(folder=folder,
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resource_pool=resource_pool,
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datastore=datastore)
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print("get_placement_spec_for_resource_pool: Result is '{}'".
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format(placement_spec))
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return placement_spec
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