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Opening files in python3 for windows10

Hi I cannot open files in python 3 actually I have a problem with the path. I don't know how to write the path for it.:/ For example I have a file(bazi.py) in folder(w8) in driver(F). How should i write it's path. Please help me im an amateur:/

In Windows, there are a couple additional ways of referencing a file. That is because natively, Windows file path employs the backslash "" instead of the slash. Python allows using both in a Windows system, but there are a couple of pitfalls to watch out for. To sum them up:
Python lets you use OS-X/Linux style slashes "/" even in Windows. Therefore, you can refer to the file as 'C:/Users/narae/Desktop/alice.txt'. RECOMMENDED.
If using backslash, because it is a special character in Python, you must remember to escape every instance: 'C:\Users\narae\Desktop\alice.txt'
Alternatively, you can prefix the entire file name string with the rawstring marker "r": r'C:\Users\narae\Desktop\alice.txt'. That way, everything in the string is interpreted as a literal character, and you don't have to escape every backslash.
File Name Shortcuts and CWD (Current Working Directory)
So, using the full directory path and file name always works; you should be using this method. However, you might have seen files called by their name only, e.g., 'alice.txt' in Python. How is it done?
The concept of Current Working Directory (CWD) is crucial here. You can think of it as the folder your Python is operating inside at the moment. So far we have been using the absolute path, which begins from the topmost directory. But if your file reference does not start from the top (e.g., 'alice.txt', 'ling1330/alice.txt'), Python assumes that it starts in the CWD (a "relative path").

using the os.path.abspath function will translate the path to a version appropriate for the operating system.
os.path.abspath(r'F:\w8\bazi.py')

Long paths in Python on Windows

I have a problem when programming in Python running under Windows. I need to work with file paths, that are longer than 256 or whatsathelimit characters.
Now, I've read basically about two solutions:
Use GetShortPathName from kernel32.dll and access the file in this way.
That is nice, but I cannot use it, since I need to use the paths in a way
shutil.rmtree(short_path)
where the short_path is a really short path (something like D:\tools\Eclipse) and the long paths appear in the directory itself (damn Eclipse plugins).
Prepend "\\\\?\\" to the path
I haven't managed to make this work in any way. The attempt to do anything this way always result in error WindowsError: [Error 123] The filename, directory name, or volume label syntax is incorrect: <path here>
So my question is: How do I make the 2nd option work? I stress that I need to use it the same way as in the example in option #1.
OR
Is there any other way?
EDIT: I need the solution to work in Python 2.7
EDIT2: The question Python long filename support broken in Windows does give the answer with the 'magic prefix' and I stated that I know it in this question. The thing I do not know is HOW do I use it. I've tried to prepend that to the path but it just failed, as I've written above.

Well it seems that, as always, I've found the answer to what's been bugging me for a week twenty minutes after I seriously ask somebody about it.
So I've found that I need to make sure two things are done correctly:
The path can contain only backslashes, no forward slashes.
If I want to do something like list a directory, I need to end the path with a backslash, otherwise Python will append /*.* to it, which is a forward slash, which is bad.
Hope at least someone will find this useful.

Let me just simplify this for anyone looking for a straight answer:
For python < 3: Path needs to be unicode, prepend string with u like u'C:\\path\\to\\file'
Path needs to start with \\\\?\\ (which is escaped into \\?\) like u'\\\\?\\C:\\path\\to\\file'
No forward slashes only backslashes: / --> \\
It has to be an absolute path; it does not work for relative paths

py 3.8.2
# Fix long path access:
import ntpath
ntpath.realpath = ntpath.abspath
# Fix long path access.
In my case, this solved the problem of running a script from a long path.
(https://developers.google.com/drive/api/v3/quickstart/python)
But this is not a universal fix.
It looks like the ntpath.realpath implementation has problems. This code replaced it with a dummy.

it works for me
import os
str1=r"C:\Users\manual\demodfadsfljdskfjslkdsjfklaj\inner-2djfklsdfjsdklfj\inner3fadsfksdfjdklsfjksdgjl\inner4dfhasdjfhsdjfskfklsjdkjfleioreirueewdsfksdmv\anotherInnerfolder4aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\5qbbbbbbbbbbbccccccccccccccccccccccccsssssssssssssssss\tmp.txt"
print(len(str1)) #346
path = os.path.abspath(str1)
if path.startswith(u"\\\\"):
path=u"\\\\?\\UNC\\"+path[2:]
else:
path=u"\\\\?\\"+path
with open(path,"r+") as f:
print(f.readline())
if you get a long path(more then 258 char) issue in windows then try this .

saving/creating a file in a relative path with python 3

I'm new to Python and i'd like to build a script (Python 3) to test electronic modules and save log files.
I'd like to save the logfiles in the following format:
201410log.txt (yearmonthlog.txt)
This is done with the code:
import os
logfile=open(time.strftime('%Y%mlog.txt'), 'a')
logfile.write('This is a test\n\n\n')
This way, every month a new log file is created.
However, i'd like the logfiles to be in in a subdirectory (\logs).
I tried approaches like
logfile=open(time.strftime('\logs\%Y%mlog.txt'), 'a')
and similar things but i couldnt get any of them to work.
I searched trough other questions on stackoverflow (for example: Relative paths in Python ) and elsewhere in the internet, but i couldnt find the right solution.
Could someone point me in the right direction?
(sorry for any mistakes/spelling errors, i'm no native english speaker)

Remove the leading backslash. It makes the path absolute. Beside that, you need to escape a backslash.
logfile = open(time.strftime('logs\\%Y%mlog.txt'), 'a')
or use r'raw string literal':
logfile = open(time.strftime(r'logs\%Y%mlog.txt'), 'a')
For your current path string literal, it does not make problem. But paths like 'a\nb' will not work because \n is interpreted as newline instead of a literal backslash and n.

Open windows explorer at a specific network location in Python

I have tried so many variants of a theme to get this explorer window open at the P:\ drive, from what my little knowledge tells me, the fact the path to the folder is anywhere but the C:\ drive means it fails (it works with C:) so perhaps the path is wrong? the code below shows some of the tries i have made but still no luck, "P:" is mapped the same on all machines.
def Open_Win_Explorer_and_Select_Dir():
import subprocess
fldrname = os.path.basename(currentproject.get())
print(fldrname)
#subprocess.Popen('c:\windows\EXPLORER.EXE', cwd=(P:/Projects 2013/)
#subbprocess.Popen('c:\\windows\\EXPLORER.EXE' cwd=('P:\\Projects_2013\\')fldrname)
#subprocess.Popen(r'C:/Windows/explorer.exe', cwd=r'//WRDBSVR/Project_Data/Projects_2013/'+fldrname)
subprocess.Popen('explorer /n, /select r"\\192.168.0.27\\Project_Data\\Projects_2013\\"'+fldrname)
#subprocess.Popen('explorer /n, /select r"P:\\Project_Data\\Projects_2013\\"'+fldrname)

well to open My pc (for windows) try:
import subprocess
subprocess.Popen('explorer ""')
"#if subprocess.Popen('explorer "{0}".format(full_path)') is struck at pc\my documents.
where full_path=os.path.join("your/path")"

Appart from the fact that Ashish Nitin Patil's answer is definitly better, as using a variable for paths is always a good idea, you have a problem with your quotes:
# This line is not correct
'explorer /n, /select r"\\192.168.0.27\\Project_Data\\Projects_2013\\"'+fldrname
# ^you start a new string without ending previous one
# this one is correct
'explorer /n, /select ' + r'\192.168.0.27\Project_Data\Projects_2013\' + fldrname
# ^first ending string start
Besides, using raw strings (r"xxx") means that \ will not escape characters, so you shall not double them. If you want to double them, you do not need prepend r.
Last remark: take care to avoid string concatenation (+) when working with paths; you should use os.path.join() instead.

Following should do the job.
import subprocess
subprocess.Popen('explorer "{0}"'.format(full_folder_path))
Update -
Tested on my system -
full_path = os.path.join("P:/Project_Data/Projects_2013/",fldrname)
print full_path # Verify that it is correct
subprocess.Popen('explorer "{0}"'.format(full_path))

Dropping a file onto a python script in windows - avoid Windows use of backslashes in the argument

I know all about how Windows uses backslashes for filenames, etc., and Unix uses forward. However, I never use backslashes with strings I create in my code. However:
When windows explorer "drops" a file onto a python script, the string it passes contains backslashes. These translate into escape sequences in the strings in the sys.argv list and then I have no way to change them after that (open to suggestions there)
Is there any way I can somehow make windows pass a literal string or ... any other way I can solve this problem?
I'd love my script to be droppable, but the only thing preventing me is windows backslashes.
EDIT:
Sorry everyone, the error was actually not the passing of the string - as someone has pointed out below, but this could still help someone else:
Make sure you use absolute path names because when the Windows shell will NOT run the script in the current directory as you would from a command line. This causes permission denied errors when attempting to write to single-part path-names that aren't absolute.

Cannot reproduce. This:
import os, sys
print sys.argv
print map(os.path.exists, sys.argv)
raw_input()
gives me this:
['D:\\workspaces\\generic\\SO_Python\\9266551.py', 'D:\\workspaces\\generic\\SO_Python\\9254991.py']
[True, True]
after dropping the second file onto the first one. Python 2.7.2 (on Windows). Can you try this code out?