I am making a program where I need permission to file paths so I am using the chmod command. I get the paths from a text document. I put the paths there and organized them by rows by using the "/n" command. The problem is when I plug the variable in the string still has a /n on it. I was wondering if there was a way to have a txt document go down a line but in a way it wont interfere.
OSError: [WinError 123] The filename, directory name, or volume label syntax is incorrect: 'C:/Users/*****/Pictures/Camera Roll\n'
I have python 3.6.2 and windows 10. I am still in the early stages of learning this language.
You want to .strip() the path before sending it to chmod:
st = 'abcd\n'
st.strip() == 'abcd' # strip() trims all whitespace
Related
I'm trying to run a python script using Sublime Text 3. I'm trying to just use Ctrl+b and type the parameters in the box that comes up, but I can't for the life of me figure out how to format the Windows file paths. I keep getting a FileNotFoundError
I've tried:
"C:\dir1\dir 2\file.ext"
"C:\\dir1\\dir 2\\file.ext"
"C:/dir1/dir 2/file.ext"
Because some of my directories have spaces in them, I'm enclosing the whole thing in double quotes no matter which slash style I try. What am I missing here? None of these works.
With the first, for example, I'm getting FileNotFoundError: [Errno 2] No such file or directory: 'C:\\dir1\\dir 2\\file.ext,' [Finished in 0.3s]
The file is most definitely there and spelled correctly.
In case it matters, I'm using docopt to parse the input parameters
Upon looping a directory to delete txt files ONLY - a message is returned indicating The System cannot find the file specified: 'File.txt'.
I've made sure the txt files that I'm attempting to delete exist in the directory I'm looping. I've also checked my code and to make sure it can see my files by printing them in a list with the print command.
import os
fileLoc = 'c:\\temp\\files'
for files in os.listdir(fileLoc):
if files.endswith('.txt'):
os.unlink(files)
Upon initial execution, I expected to see all txt files deleted except for other non-txt files. The actual result was an error message "FileNotFoundError: [WinError 2] The system cannot find the file specified: 'File.txt'.
Not sure what I'm doing wrong, any help would be appreciated.
It isn't found because the the path you intended to unlink is relative to fileLoc. In fact with your code, the effect is to unlink the file relative to the current working directory. If there were *.txt files
in the cwd then the code would have unfortunate side-effects.
Another way to look at it:
Essentially, by analogy, in the shell what you're trying to do is equivalent to this:
# first the setup
$ mkdir foo
$ touch foo/a.txt
# now your code is equvalent to:
$ rm *.txt
# won't work as intended because it removes the *.txt files in the
# current directory. In fact the bug is also that your code would unlink
# any *.txt files in the current working directory unintentionally.
# what you intended was:
$ rm foo/*.txt
The missing piece was the path to the file in question.
I'll add some editorial: The Old Bard taught us to "when in doubt, print variables". In other words, debug it. I don't see from the OP an attempt to do that. Just a thing to keep in mind.
Anyway the new code:
Revised:
import os
fileLoc = 'c:\\temp\\files'
for file in os.listdir(fileLoc):
if file.endswith('.txt'):
os.unlink(os.path.join(fileLoc,file))
The fix: os.path.join() builds a path for you from parts. One part is the directory (path) where the file exists, aka: fileLoc. The other part is the filename, aka file.
os.path.join() makes a whole valid path from them using whatever OS directory separator is appropriate for your platform.
Also, might want to glance through:
https://docs.python.org/2/library/os.path.html
What I am trying to do:
I am trying to run this line of code:
RAW_FILE="filename.txt"
os.rename("source.file", RAW_FILE)
I simply want to take a file matching source.file and rename it to something contained in the RAW_FILE variable. testing this works fine with anaconda python shell.
The problem:
When running with the script I have this error:
os.rename("source.file", RAW_FILE)
OSError: [WinError 123] The filename, directory name, or volume label syntax
is incorrect: 'source.file' -> 'filename.txt\n'
I have a print in the script and it shows the filename.txt without the '\n'
I have tried (amongst many other things):
.rstrip()
.strip()
.replace('\n', '')
My question:
How can I remove this seemly invincible '\n' and what is the cause of this?
I'm trying to read in a text file to work with Word Clouds. Here is the syntax I'm trying:
# Read the whole text.
text = open(r'C:\Users\mswitajski\Desktop\alice.txt').read()
But I keep getting the following error:
FileNotFoundError: [Errno 2] No such file or directory: 'C:\\Users\\mswitajski\\Desktop\\alice.txt'
I've triple checked the file name, tried reading it as a raw file, changed the slashes and everything but I continue to get the same error.
Well, if someone reaches up to here and still could not find the solution then here is the more pythonic way of doing the absolute path in windows.
Instead of using:
text = open(r'C:\Users\mswitajski\Desktop\alice.txt').read()
use os.sep, in conjunction of os.path.join like the following:
import os
text = open(os.path.join('C:', os.sep, 'Users', 'mswitajski', 'Desktop', 'alice.txt')).read()
Try changin the path to this"
'C:\Users\mswitajski\Desktop/alice.txt'
Sometimes windows won't find/recognize the file path when the file is specified like this
'C:\Users\mswitajski\Desktop\alice.txt'
In the answer it shows up as only one \ but you still need 2 like your previous path. The only difference is the last slash /. Hope that works.
At your text raw file (alice.txt) try delete the .txt.
The file probably is named alice.txt.txt
I face the same issue and solve it by deleted the .txt.
I had to use double slashes instead of one, because python interpreted it as a escape sequence. My final string was:
C:\\Users\\ArpitChinmay's\\AppData\\Roaming\\Code\\User\\globalStorage\\moocfi.test-my-
code\\tmcdata\\TMC workspace\\Exercises\\hy\\hy-data-analysis-with-python-
2020\\part02-e04_word_frequencies\\src\\alice.txt
However, it worked this way too,
C:\\Users\\Arpit Chinmay's\\AppData\\Roaming\\Code\\User\\globalStorage\\moocfi.test-
my-code\\tmcdata\\TMC workspace\\Exercises\\hy\\hy-data-analysis-with-python-
2020\\part02-e04_word_frequencies\\src/alice.txt
I am trying to zip a file using shell script command. I am using following command:
zip ./test/step1.zip $FILES
where $FILES contain all the input files. But I am getting a warning as follows
zip warning: name not matched: myfile.dat
and one more thing I observed that the file which is at last in the list of files in a folder has the above warning and that file is not getting zipped.
Can anyone explain me why this is happening? I am new to shell script world.
zip warning: name not matched: myfile.dat
This means the file myfile.dat does not exist.
You will get the same error if the file is a symlink pointing to a non-existent file.
As you say, whatever is the last file at the of $FILES, it will not be added to the zip along with the warning. So I think something's wrong with the way you create $FILES. Chances are there is a newline, carriage return, space, tab, or other invisible character at the end of the last filename, resulting in something that doesn't exist. Try this for example:
for f in $FILES; do echo :$f:; done
I bet the last line will be incorrect, for example:
:myfile.dat :
...or something like that instead of :myfile.dat: with no characters before the last :
UPDATE
If you say the script started working after running dos2unix on it, that confirms what everybody suspected already, that somehow there was a carriage-return at the end of your $FILES list.
od -c shows the \r carriage-return. Try echo $FILES | od -c
Another possible cause that can generate a zip warning: name not matched: error is having any of zip's environment variables set incorrectly.
From the man page:
ENVIRONMENT
The following environment variables are read and used by zip as described.
ZIPOPT
contains default options that will be used when running zip. The contents of this environment variable will get added to the command line just after the zip command.
ZIP
[Not on RISC OS and VMS] see ZIPOPT
Zip$Options
[RISC OS] see ZIPOPT
Zip$Exts
[RISC OS] contains extensions separated by a : that will cause native filenames with one of the specified extensions to be added to the zip file with basename and extension swapped.
ZIP_OPTS
[VMS] see ZIPOPT
In my case, I was using zip in a script and had the binary location in an environment variable ZIP so that we could change to a different zip binary easily without making tonnes of changes in the script.
Example:
ZIP=/usr/bin/zip
...
${ZIP} -r folder.zip folder
This is then processed as:
/usr/bin/zip /usr/bin/zip -r folder.zip folder
And generates the errors:
zip warning: name not matched: folder.zip
zip I/O error: Operation not permitted
zip error: Could not create output file (/usr/bin/zip.zip)
The first because it's now trying to add folder.zip to the archive instead of using it as the archive. The second and third because it's trying to use the file /usr/bin/zip.zip as the archive which is (fortunately) not writable by a normal user.
Note: This is a really old question, but I didn't find this answer anywhere, so I'm posting it to help future searchers (my future self included).
eebbesen hit the nail in his comment for my case (but i cannot vote for comment).
Another possible reason missed in the other comments is file exceeding the file size limit (4GB).
I converted my script for unix environment using dos2unix command and executed my script as ./myscript.sh instead bash myscript.sh.
I just discovered another potential cause for this. If the permissions of the directory/subdirectory don't allow the zip to find the file, it will report this error. Actually, if you run a chmod -R 444 on the directory, and then try to zip it, you will reproduce this error, and also have a "stored 0%" report, like this:
zip warning: name not matched: borrar/enviar
adding: borrar/ (stored 0%)
Hence, try changing the permissions of the file. If you are trying to send them through email, and those email filters (like Gmail's) invent silly filters of not sending executables, don't forget that making permissions very strict when making zip compression can be the cause of the error you are reporting, of "name not matched".
spaces are not allowed:
it would fail if there are more than one files(s) in $FILES unless you put them in loop
I also encountered this issue. In my case, the line separate is CRLF in my zip shell script which causes the problem. Using LF fixed it.