Python Path Double Backslash when opening file causing error

Question

I'm trying to open a file like this:

with open(str(script_path) + '\\description.xml', 'w+') as file:

where script_path is equal to this:

script_path = os.path.dirname(os.path.realpath(__file__)) + '\\.tmp')

When I run this I get an error that there is no such file or directory because when it tries to open the file it sees the whole path as a string, including the escape strings. Is there any way around this? Obviously .replace() won't work here as it won't replace the escape string. Hoping there is a clever way to do this within the os module?

Why are there two backslashes? If description.xml is in the same directory as the script, the relativr path is just description.xml — Ahmet
– Ahmet, Commented Jul 3, 2020 at 10:06
I think I did this so it would be in the tmp folder, although I guess I could just use '.tmp/description.xml' — HarryBilney
– HarryBilney, Commented Jul 3, 2020 at 10:12
I recommend using os.path.join to create a path instead of string concatenation. This might help avoiding this problem. — nomansland008
– nomansland008, Commented Jul 3, 2020 at 11:31

Rachit Tayal · Accepted Answer · 2020-07-10 14:22:02Z

1

Not really sure why you're adding two backslashes. You can simply create the path using a single forward slash (Linux based) or backslash (win). Something like this:

script_path = os.path.dirname(os.path.realpath(__file__)) + '/tmp/description.xml'

However, better way to achieve this would be to use os.path.join as suggested by nomansland008.

>>> import os

>>> parent_dir = "xyz"
>>> dir = "foo"
>>> file_name = "bar.txt"
>>> os.path.join(parent_dir, dir, file_name)
'xyz/foo/bar.txt'

You won't have to bother about whether the string has slash(or not). It will be taken care by join. In your case it can simply be:

os.path.join(os.path.dirname(os.path.realpath(__file__)), 'tmp', 'description.xml')

Should work, provided the files and directories exist.

answered Jul 10, 2020 at 14:22

Rachit Tayal

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