Error with back slash in python open statement

Question

Is there any way without typing r or 'double slash' to resolve this problem.because this two tricks change the type of file:

f = open('C:\Users\alireza\Desktop\exersices.python\p5.files\1.txt')

Error:

File "<ipython-input-1-243f8d6a931c>", line 1
    f = open('C:\Users\alireza\Desktop\exersices.python\p5.files\1.txt')
            ^
SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in position 2-3: truncated \UXXXXXXXX escape

Can you please indent your code properly: stackoverflow.com/editing-help and please ask your question not only in the title, this might also be helpful: stackoverflow.com/help/how-to-ask. — colidyre
– colidyre, Commented Sep 9, 2018 at 9:56
You can use / forward slash to refer to the path of the file in windows. — Praveenkumar Beedanal
– Praveenkumar Beedanal, Commented Sep 9, 2018 at 10:30
I don't think using r'...' literals or double backslashes would change the type of the file in any way. What makes you think it does? — lenz
– lenz, Commented Sep 9, 2018 at 11:14
Why is this question tagged "unicode", I don't see anything outside ASCII? It does not take any knowledge about unicode to see that the substring "sers\ali" in not a valid hexadecimal number. — Andrey Tyukin
– Andrey Tyukin, Commented Sep 9, 2018 at 12:10

Praveenkumar Beedanal · Accepted Answer · 2018-09-09 10:32:08Z

2

This will solve your issue.

f = open('C:/Users/alireza/Desktop/exersices.python/p5.files/1.txt')

answered Sep 9, 2018 at 10:32

Praveenkumar Beedanal

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1 Comment

Eryk Sun Over a year ago

Note that if long-path support (i.e. more than 260 characters) is required via the "\\\\?\\" prefix, the path must be unicode (e.g. a u"" literal in Python 2) and use backslash as the path separator. Also if it's passed as a command-line argument, a path should use backslash since many programs blindly interpret forward slash as an option/switch. You can normalize forward slashes to backslashes via os.path.normpath.

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