Python 区域：IOError：[Errno 22] 无效模式（'w'）或文件名
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Region: IOError: [Errno 22] invalid mode ('w') or filename
提问by FaerieDrgn
I'm not sure why, but for some reason, whenever I have "region" in the file name of the output file, it gives me this error:
我不知道为什么，但出于某种原因，每当我在输出文件的文件名中有“区域”时，它都会给我这个错误：
IOError: [Errno 22] invalid mode ('w') or filename: 'path\regionlog.txt'
IOError: [Errno 22] 无效模式 ('w') 或文件名: 'path\regionlog.txt'
It does this for "region.txt", "logregion.txt", etc.
它为"region.txt"、"logregion.txt"等执行此操作。
class writeTo:
def __init__(self, stdout, name):
self.stdout = stdout
self.log = file(name, 'w') #here is where it says the error occurs
output = os.path.abspath('path\regionlog.txt')
writer = writeTo(sys.stdout, output) #and here too
Why is this? I really would like to name my file "regionlog.txt" but it keeps coming up with that error. Is there a way around it?
为什么是这样？我真的很想将我的文件命名为“regionlog.txt”，但它不断出现该错误。有办法解决吗？
采纳答案by Pavel Anossov
Use forward slashes:
使用正斜杠：
'path/regionlog.txt'
Or raw strings:
或原始字符串：
r'path\regionlog.txt'
Or at least escape your backslashes:
或者至少逃避你的反斜杠：
'path\regionlog.txt'
\ris a carriage return.
\r是回车。
Another option: use os.path.joinand you won't have to worry about slashes at all:
另一种选择：使用os.path.join，您根本不必担心斜线：
output = os.path.abspath(os.path.join('path', 'regionlog.txt'))
回答by HymanChen
In C standard language, \t, \n, \rare escape characters. \tis a transverse to the next TAB position. \nis a newline and \ris a carriage return. You should use \\ror /r, and you will solve the problem!
在 C 标准语言中\t，, \n,\r是转义字符。\t是下一个 TAB 位置的横向。\n是换行符，\r是回车符。您应该使用\\r或/r，您将解决问题！
回答by Henk van der Laak
Additionaly, Python also gives this message when trying to open a file > 50 MB from a SharePoint shared drive.
此外，当尝试从 SharePoint 共享驱动器打开大于 50 MB 的文件时，Python 也会显示此消息。
回答by Roee Anuar
Another simple solution is changing the "\r" instances in the filename path to "\R"
另一个简单的解决方案是将文件名路径中的“\r”实例更改为“\R”
![Python 使用 sys.argv[1] 时出现“列表索引超出范围”](/res/img/loading.gif)