I'm trying to change directory when calling a python file in cmd but it's not working !
I tried all types of slashes & back slashes & escaping, sometimes when the code runs, the directory isn't changing and stays the same where i start the py file and sometimes the code isn't running and i have this error Error
import os
#import sys
os.chdir('%SystemRoot%/Users/%username%/AppData/Local/Google/Chrome/User Data')
os.system('cd \"%SystemRoot%\\Users\\%username%\\AppData\\Local\\Google\\Chrome\\User Data\"')
I tried to change the system variables %SystemRoot% and %username% to words like C:/ and user2 (my system root and user name ) but still not working !
Can anyone try it in his computer and tell me what to change pls ?
Thanks !
Python does not automatically expand shell variables like %SystemRoot% and %username% to their actual values, this is what caused the error you were getting. Try os.chdir(os.path.expandvars(os.path.expanduser('%SystemRoot%/Users/%username%/AppData/Local/Google/Chrome/User Data'))) for the first line as this should expand the %username% and %SystemRoot% variables to a valid path.
EDIT: Sorry, I misunderstood your question. While this will take care of the error you were getting, you cannot change the shell's working directory from your script; see comment under your question by Ulrich Eckhardt
Related
I am trying to run a batch file through python; however, it is not recognizing the path. It stops reading the path after the space between 'Practice' and 'Folder'. How can I fix this? I've tried the r and using forward and backward slashes. Any help would be awesome. Thank you!
import os
Practice = r"C:\Users\Username\Desktop\Practice Folder\Practice.bat"
os.system(Practice)
'C:\Users\Username\Desktop\Practice' is not recognized as an internal
or external command, operable program or batch file.
Change working directory to the script directory as you are using some relative redirection paths. Pushd changes current directory to any drive and can map network drives. The && chains commands and only runs the right hand command if the left hand command succeeds. %UserProfile% is a standard environmental variable which is usually better then using a fixed path of C:\Users\Username.
import os
Practice = r'pushd "%UserProfile%\Desktop\Practice Folder" && Practice.bat'
os.system(Practice)
Try using call from subprocess module.
You need to enclose the command only in double quotes.
from subprocess import call
call(r'"C:\Users\Username\Desktop\Practice Folder\Practice.bat"')
(Notice the order of placing quotes...)
This would even work with os.system() provided you take care the order of quotation marks.
from os import system
system(r'"C:\Users\Username\Desktop\Practice Folder\Practice.bat"')
This should help fix your problem.
You probably need to use two types of quotation marks e.g.
import os
Practice = r"'C:\Users\Username\Desktop\Practice Folder\Practice.bat'"
os.system(Practice)
As it is, your string does not contain quotation marks - you need to include quotation marks within your string or else Windows will think that Folder\Practice.bat is an argument to the command rather than a continuation of the file path
Try this
import os
Practice = os.path.abspath(r"C:\Users\Username\Desktop\Practice Folder\Practice.bat")
Edit:
Something like this worked for me
os.system(r'"C:\Users\Username\Desktop\Practice Folder\Practice.bat"')
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 .
It is a code to rename all the files in a given directory but it seems while running in my terminal it giving me a syntax error at the print statement. Also if I comment the statement I get an error at the if statement of main. If I remove that too I get an error at the rename_files() function call statement.
import os
def rename_files():
#Get all the files from directory
file_list = os.listdir("/Users/arpitgarg/test")
print file_list
#Rename all the files.
for file_name in file_list:
os.rename(file_name, file_name.translate(None, "0123456789")
print file_name
if __name__ == '__main__':
rename_files()
I doubt the trace back error is 'can't find the file specified', if so your py script needs to know where the files to rename; cause its not in the current working directory.
You'll have to add:
os.chdir('the exact path to files to be renamed')
before the for loop
The file_names function does not contain a properly indented statement . Neither does the if name=='main' conditional. Also, the os.rename function call is missing a closing parenthesis . Try using a IDE next time , like pyCharm. It will highlight these syntax errors to you.
When asking for help, provide us with the necessary information to help.. in this case the actual Traceback.
As stated before, the indentation is one major error. Python uses whitespace to differentiate code blocks. http://en.wikipedia.org/wiki/Python_syntax_and_semantics#Indentation
I recommend using PyCharm, as also stated before, but it is a memory hog. If running on an older computer, I would recommend using Notepad++ or PyScripter.
I made this code and it is working but only in Linux.
import subprocess as sub
sub.Popen([r"Rscript","diccionari.R"])
Where "diccionari.R" is the name of my script in R.
Error text message: System can't found the specific file.
Can somebody help me and do that it works on windows please?
Thank you.
You should probably try the slashes the other way around as how I said it earlier.
Using full path to the .r script (e.g. "C:/myfolder/diccionari.R") instead of just the script file, and using OS independent slashes.
You should specify where Rscriptis located i.e
import subprocess as sub
cmd_line = [r"C:\\Program Files\\R\\R-3.6.0\\bin\\Rscript", "diccionari.R"]
sub.Popen(cmd_line)
watch for the \\ characters
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?