如何接收python异常消息？

因此，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，应用程序中其他地方的异常日志记录代码捕获了该异常，并将其打印出来——在本例中，该代码如下所示：

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.

failed = False
try:
    ...
except Exception:
    failed = True
finally:
    if failed:
        raise Exception("Command failed")

---

关键是它打印出了mysql连接字符串。如何防止它这样做？

您可以这样更改输出：

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.

failed = False
try:
    ...
except Exception:
    failed = True
finally:
    if failed:
        raise Exception("Command failed")

使用以下语法：

raise Exception("Command failed") from None

见）：

或

if subprocess.call（）！=0:
引发异常（“命令失败”）

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

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