越过自定义异常但打印所有其他异常

Question

我是运行下面的try-except代码：

try:
    paths = file_system_client.get_paths("{0}/{1}/0/{2}/{3}/{4}".format(container_initial_folder, container_second_folder, chronological_date[0], chronological_date[1], chronological_date[2]), recursive=True)
    list_of_paths=["abfss://{0}@{1}.dfs.core.windows.net/".format(storage_container_name, storage_account_name)+path.name for path in paths if ".avro" in path.name]

except Exception as e:
    if e=="AccountIsDisabled":
        pass
    else:
        print(e)

我不想在我的 try-except 遇到错误时打印以下错误，也不想在遇到此错误时停止我的程序执行：

"(AccountIsDisabled) The specified account is disabled.
RequestId:3159a59e-d01f-0091-5f71-2ff884000000
Time:2020-05-21T13:09:03.3540242Z"

我只想超越它和打印任何其他error/exception（例如TypeError、ValueError等）会发生。

这在Python 3 中可行吗？

请注意，.get_paths() 方法属于 azure.storage.filedatalake 模块，可直接连接 Python 与 Azure Data Lake 进行路径提取。

我给出的说明是为了指出我试图绕过的异常不是内置异常。

[更新] 在按照建议的附加答案进行排序后，我将代码修改为：

import sys
from concurrent.futures import ThreadPoolExecutor
from azure.storage.filedatalake._models import StorageErrorException
from azure.storage.filedatalake import DataLakeServiceClient, DataLakeFileClient

storage_container_name="name1" #confidential
storage_account_name="name2" #confidential
storage_account_key="password" #confidential 
container_initial_folder="name3" #confidential
container_second_folder="name4" #confidential

def datalake_connector(storage_account_name, storage_account_key):
    global service_client

    datalake_client = DataLakeServiceClient(account_url="{0}://{1}.dfs.core.windows.net".format("https", storage_account_name), credential=storage_account_key)
    print("Client successfuly created!")

    return datalake_client

def create_list_paths(chronological_date, 
                      container_initial_folder="name3", 
                      container_second_folder="name4", 
                      storage_container_name="name1", 
                      storage_account_name="name2"
                      ):
    list_of_paths=list()
    print("1. success")
    paths = file_system_client.get_paths("{0}/{1}/0/{2}/{3}/{4}".format(container_initial_folder, container_second_folder, chronological_date[0], chronological_date[1], chronological_date[2]), recursive=True)
    print("2. success")
    list_of_paths=["abfss://{0}@{1}.dfs.core.windows.net/".format(storage_container_name, storage_account_name)+path.name for path in paths if ".avro" in path.name]
    print("3. success")
    list_of_paths=functools.reduce(operator.iconcat, result, [])

    return list_of_paths

service_client = datalake_connector(storage_account_name, storage_account_key)
file_system_client = service_client.get_file_system_client(file_system=storage_container_name)

try:
    list_of_paths=[]
    executor=ThreadPoolExecutor(max_workers=8)
    print("Start path extraction!")
    list_of_paths=[executor.submit(create_list_paths, i, container_initial_folder, storage_container_name, storage_account_name).result() for i in date_list]

except:
    print("no success")
    print(sys.exc_info())

不幸的是，StorageErrorException 无法处理是有原因的，我仍然得到以下标准输出：

Answer 1

你想比较异常的type，把你的条件改成：

if type(e)==AccountIsDisabled:

示例：

class AccountIsDisabled(Exception):
    pass

print("try #1")
try:
    raise AccountIsDisabled
except Exception as e:
    if type(e)==AccountIsDisabled:
        pass
    else:
        print(e)

print("try #2")
try:
    raise Exception('hi', 'there')
except Exception as e:
    if type(e)==AccountIsDisabled:
        pass
    else:
        print(e)

输出：

try #1
try #2
('hi', 'there')

Answer 2

清单[Python 3.Docs]: Compound statements - The try statement.

有几种方法可以实现这一点。这是一个：

try:
    # ...
except StorageErrorException:
    pass
except:
    print(sys.exc_info()[1])

请注意 except: 很棘手，因为您可能会默默地处理不应该处理的异常。另一种方法是捕获代码可能显式引发的任何异常。

try:
    # ...
except StorageErrorException:
    pass
except (SomeException, SomeOtherException, SomeOtherOtherException) as e:
    print(e)

快速浏览[MS.Docs]: filedatalake package and the sourcecode, revealed that StorageErrorException (which extends [MS.Docs]: HttpResponseError class)是你需要处理的

可能要检查 [SO]: About catching ANY exception。

与捕获异常失败有关，貌似有2个同名：

azure.storage.blob._generated.models._models_py3.StorageErrorException（当前导入）
azure.storage.filedatalake._generated.models._models_py3.StorageErrorException

我不知道其中的原理（我没有使用该包），但考虑到该包引发了另一个包中定义的异常，而它也定义了一个具有相同名称的包，这似乎很蹩脚。
无论如何导入正确的异常都可以解决问题。

附带说明一下，在处理这种情况时，不要只导入基本名称，而要使用完全限定名称：

import azure.storage.filedatalake._generated.models.StorageErrorException

越过自定义异常但打印所有其他异常

Overpass a custom exception but print all the other exceptions

python

exception

try-except