Best Python code snippet using localstack_python
test_lambda.py
Source:test_lambda.py
...1689 FunctionName=func_name, Qualifier=new_version["Version"]1690 )1691 e.match("ProvisionedConcurrencyConfigNotFoundException")1692 @pytest.mark.skip(reason="very slow (only execute when needed)")1693 def test_lambda_provisioned_concurrency_doesnt_apply_to_latest(1694 self, lambda_client, logs_client, create_lambda_function1695 ):1696 """create fn â publish version â provisioned concurrency @version â test if it applies to call to $LATEST"""1697 func_name = f"test_lambda_{short_uid()}"1698 create_lambda_function(1699 func_name=func_name,1700 handler_file=TEST_LAMBDA_INTROSPECT_PYTHON,1701 runtime=LAMBDA_RUNTIME_PYTHON38,1702 client=lambda_client,1703 timeout=2,1704 )1705 fn = lambda_client.get_function_configuration(FunctionName=func_name, Qualifier="$LATEST")1706 assert fn["State"] == "Active"1707 first_ver = lambda_client.publish_version(...
test_lambda_api.py
Source:test_lambda_api.py
...587class TestLambdaReservedConcurrency:588 # TODO: make this more robust & add snapshot589 @pytest.mark.skip(reason="very slow (only execute when needed)")590 @pytest.mark.aws_validated591 def test_lambda_provisioned_concurrency_doesnt_apply_to_latest(592 self, lambda_client, logs_client, create_lambda_function593 ):594 """create fn â publish version â provisioned concurrency @version â test if it applies to call to $LATEST"""595 func_name = f"test_lambda_{short_uid()}"596 create_lambda_function(597 func_name=func_name,598 handler_file=TEST_LAMBDA_INTROSPECT_PYTHON,599 runtime=Runtime.python3_8,600 client=lambda_client,601 timeout=2,602 )603 fn = lambda_client.get_function_configuration(FunctionName=func_name, Qualifier="$LATEST")604 assert fn["State"] == "Active"605 first_ver = lambda_client.publish_version(...
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