python - 返回非 json 对象作为 lambda 响应

Question

当流程未按预期工作时，我希望能够返回异常作为我的 lambda 响应，因此我正在执行类似的操作

def handler(event, context):
    try:
        project = event['environment']
    except Exception as ex:
        return {
            'statusCode': 500,
            'body': json.dumps(ex)
        }

但是这会过早失败，如下所示

{"errorMessage": "Object of type KeyError is not JSON serializable", "errorType": "TypeError", "stackTrace": ["  File \"/var/task/hr-management.py\", line 23, in handler\n    'body': json.dumps(ex)\n", "  File \"/var/lang/lib/python3.8/json/__init__.py\", line 231, in dumps\n    return _default_encoder.encode(obj)\n", "  File \"/var/lang/lib/python3.8/json/encoder.py\", line 199, in encode\n    chunks = self.iterencode(o, _one_shot=True)\n", "  File \"/var/lang/lib/python3.8/json/encoder.py\", line 257, in iterencode\n    return _iterencode(o, 0)\n", "  File \"/var/lang/lib/python3.8/json/encoder.py\", line 179, in default\n    raise TypeError(f'Object of type {o.__class__.__name__} '\n"]}%

我也尝试过:

def handler(event, context):
    try:
        project = event['environment']
    except Exception as ex:
        return {
            'statusCode': 500,
            'body': ex
        }

有什么建议吗？

Answer 1

如果您只是以字符串形式查找异常，请使用:

def handler(event, context):
    try:
        project = event['environment']
    except Exception as ex:
        return {
            'statusCode': 500,
            'body': json.dumps(ex.__repr__())
        }

___repr__() 返回一个字符串，用于格式化类，例如 print(class_example)，它是一个可读性工具。