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
Related
I first noticed this problem a bit ago when I posted this thread. Essentially, I'm getting a very strange issue where Python "sees" files that don't actually exist in my file browser when working with files.
As in, if I open a file with an absolute path, modify and print the contents of it through Python, it will print exactly what it's supposed to. But when I try to open the same file through the same absolute path on my Windows file browser, the file is not updated with my modifications.
I experienced a similar issue recently as well - renaming files. Here's a simple script I wrote to replace all spaces in filenames with underscores.
rename.py:
import os
i = 0
path = "E:/sample_path/"
for filename in os.listdir(path):
src = path + filename
dst = path + filename.replace(" ", "_")
os.rename(src, dst)
i += 1
print(str(i) + " files processed.")
Upon doing some prints in Python, I can see that all the files in the directory are being renamed correctly but it just wasn't correctly updating when I actually viewed the directory. Both in the file browser and in using the dir command. Same with creating new files in Python, they exist in the eyes of Python, but they are nowhere to be found in Windows, even with hidden files being visible and all.
Now for the interesting part: This script works if I open the python editor in cmd and import it. So I know it's all correct, no syntax errors or anything - it's just a strange error occurring with Python.
For example, if I go into command prompt and type python rename.py it won't return any errors and will even output the correct results - x files processed. but it will not actually modify any "real" files.
But if I go into command prompt and type python to bring up the cmd editor, then type import rename it gives the correct output and updates all the files correctly. So for the time being this workaround helps, but it's a very strange issue and I have yet to see anyone else encounter it. It's almost like Python creates a temporary duplicate of the filesystem and is not completing it's sync back to Windows.
I've tried reinstalling Python several times, both x64 and x86 verisons and nothing has fixed it so far.
EDIT: Here's a simplified example illustrating my issue.
write.py:
f = open("newfile.txt", "w+")
f.write("hello new file")
f.close()
read.py:
f = open("newfile.txt", "r")
for l in f.readlines():
print(l)
If I run write.py, no errors are returned. But also no file named newfile.txt is visible in my current working directory. However, if I run read.py it prints hello new file. So clearly both of these are accessing the same invisible file somewhere.
Ok so I kinda dropped the ball. I was trying to understand how things work. I had a few html files on my computer that I was trying to rename as txt files. This was strictly a learning exercise. Following the instructions I found here using this code:
for file in *.html
do
mv "$file" "${file%.html}.txt"
done
produced this error:
mv: rename *.html to *.txt: No such file or directory
Long story short I ended up going rogue and renaming the html files, as well as a lot of other non html files as txt files. So now I have files labeled like
my_movie.mp4.txt
my_song.mp3.txt
my_file.txt.txt
This may be a really dumb question but.. Is there a way to check if a file has two extensions and if yes remove the last one? Or any other way to undo this mess?
EDIT
Doing this find . -name "*.*.txt" -exec echo {} \; | cat -b seems to tell me what was changed and where it is located. The cat -b part is not necessary but I like it. This still doesn't fix what I broke though.
I'm not sure if terminal can check for extensions "twice", but you can check for . in every name an if there's more than one occurence of ., then your file has more extensions. Then you can cut the extension off with finding first occurence of . in a string when going backwards... or last one if checking characters in string in a normal way.
I have a faster option for you if you can use python. You can strip the extension with:
for file in list_of_files:
os.rename(file,os.path.splitext(file)[0])
which can give you from your file.txt.txt your file.txt
Example:
You wrote that your command tells you what has changed, so just take those changed files and dump them into a file(path to file per line). Then you can easily run this:
with open('<path to list>') as f:
list_of_files = f.readlines()
for file in list_of_files:
os.rename(file.strip('\n'), os.path.splitext(file.strip('\n'))[0])
If not, then you'd need to get the list from python:
import os
results = []
for root, folder, filenames in os.walk(<your path to folder>):
for filename in filenames:
if filename.endswith('.txt.txt'):
results.append(os.path.join(root, filename))
With this you got a list of files ending with .txt.txt like this <your folder>\\<path_to_file>.
Get a path to your directory used in os.walk() without folder's name(it's already in list) so it'll be like this:
e.g. os.walk('/home/me/directory') -> path='/home/me/' and res is item already in a list, which looks like directory/...
for res in results:
path = '' # set the path here
file = os.path.join(path,r)
os.rename(file, os.path.splitext(file)[0])
Depending on what files you want to find change .txt.txt in if filename.endswith('...') to whatever you like and os.rename() will take file's name without extension which in your case means it strips the additional extension you don't want to have.
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.
I'm using ImageMagick to do some image processing from the commandline, and would like to operate on a list of files as specified in foo.txt. From the instructions here: http://www.imagemagick.org/script/command-line-processing.php I see that I can use Filename References from a file prefixed with #. When I run something like:
montage #foo.txt output.jpg
everything works as expected, as long as foo.txt is in the current directory. However, when I try to access bar.txt in a different directory by running:
montage /some_directory/#bar.txt
output2.jpg
I get:
montage: unable to open image
/some_directory/#bar.txt: No such file
or directory # blob.c/OpenBlob/2480.
I believe the issue is my syntax, but I'm not sure what to change it to. Any help would be appreciated.
Quite an old entry but it seems relatively obvious that you need to put the # before the full path:
montage #/some_directory/bar.txt output2.jpg
As of ImageMagick 6.5.4-7 2014-02-10, paths are not supported with # syntax. The # file must be in the current directory and identified by name only.
I haven't tried directing IM to pull the list of files from a file, but I do specify multiple files on the command line like this:
gm -sOutputFile=dest.ext -f file1.ppm file2.ppm file3.ppm
Can you pull the contents of that file into a variable, and then let the shell expand that variable?
There exists a function that is part of a software package (MRICro), and it is called 'dcm2nii.'
When a relative path is used as the output directory, the function works perfectly well.
But, when the absolute path to the exact same folder is used, the function breaks down.
Example (absolute path):
dcm2nii -o /net/user1/project_name/Data/2011_01_10_SVD1/Processed/3_fMRI_rest E2538S3I00*
Example (relative path):
dcm2nii -o ../Processed/3_fMRI_rest E2538S3I00*
Sample error message that occurs when using the absolute path for the output folder (the last line suggests that the output file can not be created):
Validating 52 potential DICOM images.
Found 52 DICOM images.
Converting 52/52 2
E2538S3I0001.MR.dcm->20110110_112950E2538S3I0001MRFPSD1F29OCT2010RCs003a1001.nii
GZip 20110110_112950E2538S3I0001MRFPSD1F29OCT2010RCs003a1001.nii
unable to create /net/user1/project_name/Data/2011_01_10_SVD1/Processed/3_fMRI_rest/20110110_112950E2538S3I0001MRFPSD1F29OCT2010RCs003a1001.nii.gz
I do not know if this problem is due to me doing something wrong in Linux/bash or due to the function actually having a mistake.
But, any input is appreciated.
On a more general level, I am looking for any foreseeable reason for why a function would be able to use a relative folder path and not an absolute one (provided that they resolve to the same location).
EDIT: pwd gives:
/net/user1/project_name/Data/2011_01_10_SVD1/3_fMRI_rest
you would really have to show us the code before we can tell you what the cause of the problem is, however the behaviour you described is possible
This is an example of poor practice but consider the following
#!/bin/bash
....
current_dir = $(pwd)
out_dir = $1
cd ${somewhere}
..... do stuff
#no we want to come back to create the output dir
mkdir ${current_dir}/$out_dir
This appears to be a bug of some kind.. I am experiencing the same issue. If I try to execute this command on data in my home folder, I get the same error.
However, if I move my data to a path that doesn't involve any expansion, i.e. ''/tmp/data'', the program executes fine.