重置当前的异常链

Question

我在我的应用中有这个代码：

    //Picked user.
    for (UserList selected : selectedUsers) {
        GroupMembers member = new GroupMembers();

        member.setName(selected.getName());
        member.setUsername(selected.getUsername());
        member.setEmail(selected.getEmail());
        member.setUid(selected.getUid());

        mGroupMembers.add(member);

        mPickedText.setText("Selected: " + selected.getName());
    }

当我的MongoDB出现故障时，我有这个：

try:
    return self.wsgi_app(environ, start_response)
except Exception:
    logger.warning('Failed to render the custom %s page', exc.code, exc_info=True)

我不想要有关第一个异常的信息 - 我之前已经记录了我的代码。我只想记录在try / except块内发生的异常（在处理上述异常时出现的另一个异常，发生了另一个异常：）。

如何在try / except块之前重置当前的异常链？

我可以试试：

18:44:55 WARNING Failed to render the custom 500 page
Traceback (most recent call last):
...
  File ".../venv/lib/python3.5/site-packages/pymongo/topology.py", line 189, in select_servers
    self._error_message(selector))
pymongo.errors.ServerSelectionTimeoutError: localhost:27017: [Errno 61] Connection refused

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
...
pymongo.errors.ServerSelectionTimeoutError: localhost:27017: [Errno 61] Connection refused

但它会破坏可能在try / except中发生的几个例外的有用链。