尝试读取文件的Python方法，并在出现异常时回退到备用文件

Question

尝试读取文件的Python方法是什么？如果此读取引发异常回退以读取备用文件？

这是我编写的示例代码，它使用嵌套的try-except块。这是蟒蛇吗：

try:
with open(file1, "r") as f:
params = json.load(f)
except IOError:
try:
with open(file2, "r") as f:
params = json.load(f)
except Exception as exc:
print("Error reading config file {}: {}".format(file2, str(exc)))
params = {}
except Exception as exc:
print("Error reading config file {}: {}".format(file1, str(exc)))
params = {}

Answer 1

对于两个文件，我认为这种方法已经足够好了。

如果你有更多的文件要备份，我会使用一个循环：

for filename in (file1, file2):
try:
with open(filename, "r") as fin:
params = json.load(f)
break
except IOError:
pass
except Exception as exc:
print("Error reading config file {}: {}".format(filename, str(exc)))
params = {}
break
else:   # else is executed if the loop wasn't terminated by break
print("Couldn't open any file")
params = {}

Answer 2

您可以先检查file1是否存在，然后决定打开哪个文件。它将缩短代码并避免重复try -- catch子句。我认为它更像Python，但请注意，您需要在模块中使用import os才能正常工作。它可以类似于：

fp = file1 if os.path.isfile(file1) else file2
if os.path.isfile(fp):
try:
with open(fp, "r") as f:
params = json.load(f)
except Exception as exc:
print("Error reading config file {}: {}".format(fp, str(exc)))
params = {}
else:
print 'no config file'

Answer 3

虽然我不确定这是否是蟒蛇，但可能是类似于：

file_to_open = file1 if os.path.isfile(file1) else file2

Answer 4

我建议使用pathlib.Path。

from pathlib import Path
path1 = Path(file1)
path2 = Path(file2)
path = path1 if path1.exists() else path2
with path.open('r') as f:
params = json.load(f)

如果需要，您可以添加额外的错误检查path2是否存在。