Merge "Fix show instance with integer name NotFound"
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commit
0cbdfd5fcd
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@ -375,6 +375,10 @@ class FindResourceTestCase(testtools.TestCase):
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output = utils.find_resource(self.manager, 'entity_three')
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self.assertEqual(self.manager.get('4242'), output)
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def test_find_by_int_name(self):
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output = utils.find_resource(self.manager, 9876)
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self.assertEqual(self.manager.get('5678'), output)
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class ResourceTest(testtools.TestCase):
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def setUp(self):
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@ -214,10 +214,12 @@ def find_resource(manager, name_or_id):
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"""Helper for the _find_* methods."""
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# first try to get entity as integer id
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# if the 'id' starts with '0' don't treat it as an int
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if isinstance(name_or_id, int) or (
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name_or_id.isdigit() and not name_or_id.startswith('0')):
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name_or_id = int(name_or_id)
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# When the 'name_or_id' is int, covert it to string.
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# Reason is that manager cannot find instance when name_or_id
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# is integer and instance name is digital.
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# Related to bug/1740015.
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if isinstance(name_or_id, int):
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name_or_id = six.text_type(name_or_id)
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elif sys.version_info <= (3, 0):
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name_or_id = encodeutils.safe_decode(name_or_id)
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