The program is supposed to find a file and return whether it exists on the system or not, now i have found that find command should be used, but since this command will be initiated by the code (using System) i need to save the results in a file, and trying on the terminal, i cannot get it to work, the result doesn't appear in the file, i am trying:
find / -name 'test2abc' 2>/dev/null -> res
The file res is empty. How to do it right?
Also is there a better way to do it, i am supposed to print the details of the file using stat command if the file exists. Is there a way to use juts the stat command to search for the file in the subfolders as well?
The -> res part should be > res only.
If you try the command like this on the commandline:
find / -name 'test2abc' -> res
it will print an error:
find: paths must precede expression: -
The - is not part of any valid redirection and hence given to find which cannot interpret it either.
It may be wise not to suppress error messages. A simple way can be redirecting both stderr and stdout to the file like this:
find / -name 'test2abc' > res 2>&1
Then the error about the - would have been in the file right from the start you you would have known what`s wrong very fast.
Related
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
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 have been working on this for quite some time and decided to ask for some help. I'm trying to use a command to find a multiple occurrences of a function (basically a string) within a directory (that has multiple files) and would like to view only the file names which the string is found.
Lets say this was the directory I want to search filled with multiple .h and .cpp files is:
~/Project/Files
and I was looking for occurrences of a function called 'doThis'
So far I have tried:
grep -r doThis ~/Project/Files
But I get the path and where it occurs in the file, I only need the file names.
Also grep -f wont work because I get an error message saying "No such file or directory" and when using just grep I get an error message saying "path is a directory"
Any help would be great: Thanks guys!
Simply use the -l switch ;)
So :
grep -rl foobar dir
I'm trying to redirect(?) my standard error/output to a text file.
I did my research, but for some reason the online answers are not working for me.
What am I doing wrong?
cd /home/user1/lists/
for dir in $(ls)
do
(
echo | $dir > /root/user1/$dir" "log.txt
) > /root/Desktop/Logs/Update.log
done
I also tried
2> /root/Desktop/Logs/Update.log
1> /root/Desktop/Logs/Update.log
&> /root/Desktop/Logs/Update.log
None of these work for me :(
Help please!
Try this for the basics:
echo hello >> log.txt 2>&1
Could be read as: echo the word hello, redirecting and appending STDOUT to the file log.txt. STDERR (file descriptor 2) is redirected to wherever STDOUT is being pointed. Note that STDOUT is the default and thus there is no "1" in front of the ">>". Works on the current line only.
To redirect and append all output and error of all commands in a script, put this line near the top. It will be in effect for the length of the script instead of doing it on each line:
exec >>log.txt 2>&1
If you are trying to obtain a list of the files in /home/user1/lists, you do not need a loop at all:
ls /home/usr1/lists/ >Update.log
If you are attempting to run every file in the directory as an executable with a newline as its input, and collect the output from all these programs in Update.log, try this:
for file in /home/user1/lists/*; do
echo | "$file"
done >Update.log
(Notice how we avoid the useless use of ls and how there is no redirection inside the loop.)
If you want to create an empty file called *.log.txt for each file in the directory, you would do
for file in /home/user1/lists/*; do
touch "$(basename "$file")"log.txt
done
(Using basename to obtain the file name without the directory part avoids the cd but you could do it the other way around. Generally, we tend to avoid changing the directory in scripts, so that the tool can be run from anywhere and generate output in the current directory.)
If you want to create a file containing a single newline, regardless of whether it already exists or not,
for file in /home/user1/lists/*; do
echo >"$(basename "$file")"log.txt
done
In your original program, you redirect the echo inside the loop, which means that the redirection after done will not receive any output at all, so the created file will be empty.
These are somewhat wild guesses at what you might actually be trying to accomplish, but should hopefully help nudge you slightly in the right direction. (This should properly be a comment, I suppose, but it's way too long and complex.)
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.