python - 我怎样才能吞掉Python异常消息？

Question

我有一些代码:

try:
    subprocess.check_call(
        "mysqldump {} {}".format(mysql_args, cmd), shell=True
    )
except Exception as e:
    raise Exception("Command failed")

问题在于应用程序中其他位置的异常日志记录代码捕获此异常并有效地将其打印出来 - 在本例中如下所示:

Traceback (most recent call last):
  File "/apps/django/myapp/tasks.py", line 336, in _safe_mysqldump
    "mysqldump {} {}".format(mysql_args, cmd), shell=True
  File "/usr/lib/python3.6/subprocess.py", line 291, in check_call
    raise CalledProcessError(retcode, cmd)
subprocess.CalledProcessError: 
Command 'mysqldump -ufoo -pMYSECRETPASSWORD myappdb t1 t2 t3...' returned non-zero exit status 6.

During handling of the above exception, another exception occurred:

etc.

关键是它打印出了 mysql 连接字符串。我怎样才能阻止它这样做？

Answer 1

使用语法:

raise Exception("Command failed") from None

参见PEP 409 Suppressing exception context ):

>>> try:
...     raise TypeError('a')
... except TypeError:
...     raise ValueError('b') from None
... 
Traceback (most recent call last):
  File "<stdin>", line 4, in <module>
ValueError: b

与您看到的默认行为相比:

>>> try:
...     raise TypeError('a')
... except TypeError:
...     raise ValueError('b')
... 
Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
TypeError: a

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "<stdin>", line 4, in <module>
ValueError: b

<小时/>

作为特定于您的用例的替代方案:不要使用会引发异常的 check_call ，而是使用 subprocess.run 相反:

if subprocess.run(<command>).returncode != 0:
    raise Exception("Command failed")

或subprocess.call

if subprocess.call(<command>) != 0:
    raise Exception("Command failed")