python - 我应该如何在 Python 中捕获 `with open(filename)` 引发的异常？

Question

尝试在 Python 中打开文件的行为可能会引发异常。如果我使用 with 语句打开文件，是否可以捕获 open 调用和相关的 __enter__ 调用抛出的异常而不捕获异常由 with block 中的代码引发？

try:
    with open("some_file.txt") as infile:
        print("Pretend I have a bunch of lines here and don't want the `except` below to apply to them")
        # ...a bunch more lines of code here...
except IOError as ioe:
    print("Error opening the file:", ioe)
    # ...do something smart here...

这个问题不同于this older one因为旧的是关于编写上下文管理器，而不是使用熟悉的 with open。

Answer 1

can I catch exceptions thrown by the open call and the related __enter__ call without catching exceptions raised by the code within the with block?

是的:

#!/usr/bin/env python3
import contextlib

stack = contextlib.ExitStack()
try:
    file = stack.enter_context(open('filename'))
except OSError as e:
    print('open() or file.__enter__() failed', e)
else:
    with stack:
        print('put your with-block here')

使用默认的 open() 函数，__enter__() 不应引发任何有趣的异常，因此可以简化代码:

#!/usr/bin/env python    
try:
    file = open('filename')
except OSError as e:
    print('open() failed', e)
else:
    with file:
        print('put your with-block here')