Python:为什么要打印回溯？

Question

我有个功能。函数可以在更多的线程中启动。我试着打印自己的错误信息。但我所做的并不重要--我还是会把追踪印出来的。我的职能：

def getSuggestengineResultForThree(suggestengine, seed, dynamoDBLocation):
    results[seed][suggestengine] = getsuggestsforsearchengine(seed, suggestengine)

    for keyword_result in results[seed][suggestengine]:
        o = 0
        while True:
            try:
                allKeywords.put_item(
                    Item={
                        'keyword': keyword_result
                    }
                )
                break
            except ProvisionedThroughputExceededException as pe:
                if (o > 9):
                    addtoerrortable(keyword_result)
                    print('ProvisionedThroughputExceededException 10 times in getSuggestengineResultForThree for allKeywords')
                    break
                sleep(1)
                o = o + 1
                print("ProvisionedThroughputExceededException in getSugestengineResult")

但是每一个线程我都会得到这样的输出：

Exception in thread Thread-562:
Traceback (most recent call last):
  File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/threading.py", line 914, in _bootstrap_inner
self.run()
  File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/threading.py", line 862, in run
self._target(*self._args, **self._kwargs)
  File "/Users/iTom/ownCloud/Documents/Workspace/PyCharm/Keywords/TesterWithDB.py", line 362, in getSuggestengineResultForThree
'keyword': keyword_result
  File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/boto3/resources/factory.py", line 518, in do_action
response = action(self, *args, **kwargs)
  File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/boto3/resources/action.py", line 83, in __call__
response = getattr(parent.meta.client, operation_name)(**params)
  File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/botocore/client.py", line 252, in _api_call
return self._make_api_call(operation_name, kwargs)
  File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/botocore/client.py", line 542, in _make_api_call
raise ClientError(parsed_response, operation_name)
botocore.exceptions.ClientError: An error occurred (ProvisionedThroughputExceededException) when calling the PutItem operation: The level of configured provisioned throughput for the table was exceeded. Consider increasing your provisioning level with the UpdateTable API

有人能帮我得到我自己的指纹而不是回溯吗？

Answer 1

可能ProvisionedThroughputExceededException实际上不是错误。尝试：

except botocore.exceptions.ClientError as pe:

而不是。

如果这不起作用，就找出错误发生在哪一行，并将except语句放在那里。

Answer 2

这个答案对你的问题有点迟，但这里是为了防止有人在寻找这个答案。

尼尔回答说，boto3抛出的例外是一个botocore.exceptions.ClientError。但是，您应该检查“ProvisionedThroughputExceededException”的响应错误代码，因为ClientError可能是另一个问题。

from botocore.exceptions import ClientError

except ClientError as e:
  if e.response['Error']['Code'] != 'ProvisionedThroughputExceededException':
    raise
  # do something else with 'e'

我正在使用Python2.7，这可能会或不会产生影响。我收到的异常表明，boto3已经在进行重试(最多9次)，这与您的异常不同：

调用ProvisionedThroughputExceededException操作时发生错误(达到最大值重试: 9)：超出了表的配置配置吞吐量级别。考虑使用UpdateTable API来提高您的配置级别。