Bash script not deleting files in given directory - linux

I found this bash script online that I want to use to delete files older than 2 days:
#!/bin/bash
find /path/to/dir -type f -mtime +2 -exec rm {} \;
I setup a cronjob to run the script (I set it a couple of minutes ahead for testing, but it should run once every 24 hours)
54 18 * * * /path/to/another/dir/script.sh
I exit correct so it updates the cronjob.
Why does it not delete the files in the directory?

What if you try dumping an echo at the end of the script and log the output
cron1.sh >> /var/log/cron1.log
You could try this but I'm not sure it will work
--exec rm -rf {}

Most cron jobs do not have PATH set. You must fully qualify the find command.
#!/bin/bash
/usr/bin/find /path/to/dir -type f -mtime +2 -exec rm {} \;
If you capture the stdout and stderr as recommended by damienfrancois, you'd probably see the message "command not found: find". If you didn't capture the stdout and stderr, cron usually will send the output to the cron job owner's email, unless configured not to do so.

Related

Cron Job email find: missing argument to `-exec'

I'm setting up a webserver with Plesk on Ubuntu 18.04 and I would like to use a part of the space I've available to store security footage. I succeeded in automatically uploading the photos and videos to the correct folder, but the problem is that they are not automatically removed, so that the server is full of security images. I upload the footages to a folder on the server that is also available from the internet (secured). I did some research on the internet to a cron job that automatically deleted the files older than 7 days where I found this:
find /var/www/vhosts/path to files/* -mtime +7 -exec rm -f {} \;
I also found that you can name a file to, for example: delete-files and which can be executed with crontab -e. (Yes, I made it executable;-)
I added this cron to run every hour and stated that I received notifications from the cron. Now, however, I get the following output: find: missing argument to `-exec '
Is there anything else that I need to share? Like logs?
change find /var/www/vhosts/path to files/* -mtime +7 -exec rm -f {} \;
to
find /var/www/vhosts/path to files/ -mtime +7 -exec rm -f {} \;
the * is unnecessary in the path
Can you try this as well?
find /var/www/vhosts/path to files/ -mtime +7 | xargs rm -f

Schedule cronjob to remove files modified before X days

I want to create and schedule cronjob to remove files modified before x days
I have taken following steps for that
Created Shell script (Named: Script.sh) as bellow
#!/bin/sh
15 2 * * 2-6 find /usr/sch/cbm/files/newui/log -type f -mtime +2 exec rm {} \;
I have put this file in "/var/spool/cron/crontabs" and usr/bin folder because i wasn't sure where to place exactly.
When i check with Crontab -e command get as in bellow image
But i didn't found any effect on my files. I am not sure is my job scheduled or i required to do anything else still
please guide me
crontab <file>
where file is what you have above will put the file in the right place and make sure everything is happy for cron to run it.
crontab -l
will display the list of cron jobs for the currently logged in user.
Note that the file itself does not execute so no need for #!/bin/sh as the first line. It is just a data file that cron interprets.
man 5 crontab
and
man 1 crontab
for more information.
You have to insert cron job line in cron file of user who executes that job.
su - username
crontab -e
add cron job line
save and exit
Example for removing files before x days
00 00 * * * find /path/to/folder -mtime +x -exec rm {} \;
or you can do that for some files exist in folder
For example remove files end with .log
00 00 * * * find /path/to/folder/*.log -mtime +x -exec rm {} \;

Remove log files using cron job

Hi. I want to remove all log files from the last 7 days from a folder, but leave all the other files. Can I use the below command? How do you specify that it just delete the files with .log extension?
find /path/to/file -mtime +7 -exec rm -f {} \;
Do I need to write this command into some file, or can I just write it in command prompt and have it run automatically every day?
I have no idea how to run a cron job in linux.
Use wildcard. And just put it in your crontab use the crontab -e option to edit your crontab jobs.
See example:
* * * * * find /path/to/*.log -mtime +7 -exec rm -f {} \;
Just to increment the answer check this nice article on how to work with your crontab ! in Linux .
You edit your personal crontab by running crontab -e.
This gets saved to /var/spool/cron/<username>. The file will be the owners username, so root would be /var/spool/cron/root. Everything in the file is run as the owner of the file.
The syntax for crontab is as follows:
SHELL=/bin/bash
PATH=/sbin:/bin:/usr/sbin:/usr/bin
MAILTO=root
HOME=/
# For details see man 4 crontabs
# Example of job definition:
# .---------------- minute (0 - 59)
# | .------------- hour (0 - 23)
# | | .---------- day of month (1 - 31)
# | | | .------- month (1 - 12) OR jan,feb,mar,apr ...
# | | | | .---- day of week (0 - 6) (Sunday=0 or 7) OR sun,mon,tue,wed,thu,fri,sat
# | | | | |
# * * * * * user-name command to be executed
When you are editing your own personal crontab, via crontab -e, you leave out the user-name field, because the user is inferred by the filename (see first paragraph).
That being said, your entry should look like this:
0 5 * * * find /path/to/*.log -mtime +7 -delete
This will run every day, at 5:00 AM, system time. I don't think you need it to run any more frequently than daily, given the fact that you are removing files that are 7 days old.
Please don't use over use the -exec option, when the -delete option does exactly what you want to do. The exec forks a shell for every file, and is excessively wasteful on system resources.
When you are done, you can use crontab -l to list your personal crontab.
ps. The default editor on most Linux systems is vi, if you do not know vi, use something simple like nano by setting your environ variable export EDITOR=nano
find /path/to/dir-containing-files -name '*.log' -mtime +7 -exec rm -f {} \;
To create a cron job, put a file containing the following in the /etc/cron.daily dir:
#!/bin/sh
find /path/to/dir-containing-files -name '*.log' -mtime +7 -exec rm -f {} \;
You should use crontab -e to edit your crontab and schedule the job. It might look something like this:
* 1 * * * /usr/bin/find /path/to/file -name '*.log' -mtime +7 -exec rm -f {} \;
This will recursively remove all .log files in the directory /path/to/file every day at 1am.
Since this is about log files, you should look at logrotate. It runs daily from system cron job and will rotate logs for you based on rules from /etc/logrotate.conf file, which usually includes /etc/logrotate.d directory. So no need for crontab nor find.
You can also have your own cron job if you have no access to add file to /etc/logrotate.d for your own configuration.
There are plenty of examples in /etc/logrotate.d.
It expects your application to write to single file. It is not for an application that logs into different log file each day. An application generally needs not do that. If the application keeps the log file open, logrotate can run a postrotate script to tell the application to reopen the log file.
You guys are doing it the HARD way. Try using the clear command
* * * * 0 clear > /home/user/CronLog.txt:
where 0 is Sunday and 7 would be Saturday. the ">" will clear the log as appose to ">>" which adds to the log. If your log file is root then type in "root" before "clear" like this
* * * * 0 root clear > /home/user/CronLog.txt
After googling around on this particular topic, I found that many people recommend using the -delete option like so:
* * * * * find /path/to/*.log -mtime +7 -delete;
The benefits of this version is that it is easy to remember and it will perform better since -exec will spawn a new process for every file that is to be deleted.
Here are some references:
https://linuxaria.com/howto/linux-shell-how-to-use-the-exec-option-in-find-with-examples
https://unix.stackexchange.com/questions/167823/find-exec-rm-vs-delete
This will delete log files older than 7 Days
* * * * * find /path/to -name '*.log' -mtime +7 -exec rm -f {} \;
This will delete log files older than 30 Minutes
* * * * * find /path/to -name '*.log' -mmin +30 -exec rm -f {} \;

Argument list too long error for rm, cp, mv commands

I have several hundred PDFs under a directory in UNIX. The names of the PDFs are really long (approx. 60 chars).
When I try to delete all PDFs together using the following command:
rm -f *.pdf
I get the following error:
/bin/rm: cannot execute [Argument list too long]
What is the solution to this error?
Does this error occur for mv and cp commands as well? If yes, how to solve for these commands?
The reason this occurs is because bash actually expands the asterisk to every matching file, producing a very long command line.
Try this:
find . -name "*.pdf" -print0 | xargs -0 rm
Warning: this is a recursive search and will find (and delete) files in subdirectories as well. Tack on -f to the rm command only if you are sure you don't want confirmation.
You can do the following to make the command non-recursive:
find . -maxdepth 1 -name "*.pdf" -print0 | xargs -0 rm
Another option is to use find's -delete flag:
find . -name "*.pdf" -delete
tl;dr
It's a kernel limitation on the size of the command line argument. Use a for loop instead.
Origin of problem
This is a system issue, related to execve and ARG_MAX constant. There is plenty of documentation about that (see man execve, debian's wiki, ARG_MAX details).
Basically, the expansion produce a command (with its parameters) that exceeds the ARG_MAX limit.
On kernel 2.6.23, the limit was set at 128 kB. This constant has been increased and you can get its value by executing:
getconf ARG_MAX
# 2097152 # on 3.5.0-40-generic
Solution: Using for Loop
Use a for loop as it's recommended on BashFAQ/095 and there is no limit except for RAM/memory space:
Dry run to ascertain it will delete what you expect:
for f in *.pdf; do echo rm "$f"; done
And execute it:
for f in *.pdf; do rm "$f"; done
Also this is a portable approach as glob have strong and consistant behavior among shells (part of POSIX spec).
Note: As noted by several comments, this is indeed slower but more maintainable as it can adapt more complex scenarios, e.g. where one want to do more than just one action.
Solution: Using find
If you insist, you can use find but really don't use xargs as it "is dangerous (broken, exploitable, etc.) when reading non-NUL-delimited input":
find . -maxdepth 1 -name '*.pdf' -delete
Using -maxdepth 1 ... -delete instead of -exec rm {} + allows find to simply execute the required system calls itself without using an external process, hence faster (thanks to #chepner comment).
References
I'm getting "Argument list too long". How can I process a large list in chunks? # wooledge
execve(2) - Linux man page (search for ARG_MAX) ;
Error: Argument list too long # Debian's wiki ;
Why do I get “/bin/sh: Argument list too long” when passing quoted arguments? # SuperUser
find has a -delete action:
find . -maxdepth 1 -name '*.pdf' -delete
Another answer is to force xargs to process the commands in batches. For instance to delete the files 100 at a time, cd into the directory and run this:
echo *.pdf | xargs -n 100 rm
If you’re trying to delete a very large number of files at one time (I deleted a directory with 485,000+ today), you will probably run into this error:
/bin/rm: Argument list too long.
The problem is that when you type something like rm -rf *, the * is replaced with a list of every matching file, like “rm -rf file1 file2 file3 file4” and so on. There is a relatively small buffer of memory allocated to storing this list of arguments and if it is filled up, the shell will not execute the program.
To get around this problem, a lot of people will use the find command to find every file and pass them one-by-one to the “rm” command like this:
find . -type f -exec rm -v {} \;
My problem is that I needed to delete 500,000 files and it was taking way too long.
I stumbled upon a much faster way of deleting files – the “find” command has a “-delete” flag built right in! Here’s what I ended up using:
find . -type f -delete
Using this method, I was deleting files at a rate of about 2000 files/second – much faster!
You can also show the filenames as you’re deleting them:
find . -type f -print -delete
…or even show how many files will be deleted, then time how long it takes to delete them:
root#devel# ls -1 | wc -l && time find . -type f -delete
100000
real 0m3.660s
user 0m0.036s
sys 0m0.552s
Or you can try:
find . -name '*.pdf' -exec rm -f {} \;
you can try this:
for f in *.pdf
do
rm "$f"
done
EDIT:
ThiefMaster comment suggest me not to disclose such dangerous practice to young shell's jedis, so I'll add a more "safer" version (for the sake of preserving things when someone has a "-rf . ..pdf" file)
echo "# Whooooo" > /tmp/dummy.sh
for f in '*.pdf'
do
echo "rm -i \"$f\""
done >> /tmp/dummy.sh
After running the above, just open the /tmp/dummy.sh file in your favorite editor and check every single line for dangerous filenames, commenting them out if found.
Then copy the dummy.sh script in your working dir and run it.
All this for security reasons.
For somone who doesn't have time.
Run the following command on terminal.
ulimit -S -s unlimited
Then perform cp/mv/rm operation.
I'm surprised there are no ulimit answers here. Every time I have this problem I end up here or here. I understand this solution has limitations but ulimit -s 65536 seems to often do the trick for me.
You could use a bash array:
files=(*.pdf)
for((I=0;I<${#files[#]};I+=1000)); do
rm -f "${files[#]:I:1000}"
done
This way it will erase in batches of 1000 files per step.
you can use this commend
find -name "*.pdf" -delete
The rm command has a limitation of files which you can remove simultaneous.
One possibility you can remove them using multiple times the rm command bases on your file patterns, like:
rm -f A*.pdf
rm -f B*.pdf
rm -f C*.pdf
...
rm -f *.pdf
You can also remove them through the find command:
find . -name "*.pdf" -exec rm {} \;
If they are filenames with spaces or special characters, use:
find -name "*.pdf" -delete
For files in current directory only:
find -maxdepth 1 -name '*.pdf' -delete
This sentence search all files in the current directory (-maxdepth 1) with extension pdf (-name '*.pdf'), and then, delete.
i was facing same problem while copying form source directory to destination
source directory had files ~3 lakcs
i used cp with option -r and it's worked for me
cp -r abc/ def/
it will copy all files from abc to def without giving warning of Argument list too long
Try this also If you wanna delete above 30/90 days (+) or else below 30/90(-) days files/folders then you can use the below ex commands
Ex: For 90days excludes above after 90days files/folders deletes, it means 91,92....100 days
find <path> -type f -mtime +90 -exec rm -rf {} \;
Ex: For only latest 30days files that you wanna delete then use the below command (-)
find <path> -type f -mtime -30 -exec rm -rf {} \;
If you wanna giz the files for more than 2 days files
find <path> -type f -mtime +2 -exec gzip {} \;
If you wanna see the files/folders only from past one month .
Ex:
find <path> -type f -mtime -30 -exec ls -lrt {} \;
Above 30days more only then list the files/folders
Ex:
find <path> -type f -mtime +30 -exec ls -lrt {} \;
find /opt/app/logs -type f -mtime +30 -exec ls -lrt {} \;
And another one:
cd /path/to/pdf
printf "%s\0" *.[Pp][Dd][Ff] | xargs -0 rm
printf is a shell builtin, and as far as I know it's always been as such. Now given that printf is not a shell command (but a builtin), it's not subject to "argument list too long ..." fatal error.
So we can safely use it with shell globbing patterns such as *.[Pp][Dd][Ff], then we pipe its output to remove (rm) command, through xargs, which makes sure it fits enough file names in the command line so as not to fail the rm command, which is a shell command.
The \0 in printf serves as a null separator for the file names wich are then processed by xargs command, using it (-0) as a separator, so rm does not fail when there are white spaces or other special characters in the file names.
Argument list too long
As this question title for cp, mv and rm, but answer stand mostly for rm.
Un*x commands
Read carefully command's man page!
For cp and mv, there is a -t switch, for target:
find . -type f -name '*.pdf' -exec cp -ait "/path to target" {} +
and
find . -type f -name '*.pdf' -exec mv -t "/path to target" {} +
Script way
There is an overall workaroung used in bash script:
#!/bin/bash
folder=( "/path to folder" "/path to anther folder" )
if [ "$1" != "--run" ] ;then
exec find "${folder[#]}" -type f -name '*.pdf' -exec $0 --run {} +
exit 0;
fi
shift
for file ;do
printf "Doing something with '%s'.\n" "$file"
done
What about a shorter and more reliable one?
for i in **/*.pdf; do rm "$i"; done
I had the same problem with a folder full of temporary images that was growing day by day and this command helped me to clear the folder
find . -name "*.png" -mtime +50 -exec rm {} \;
The difference with the other commands is the mtime parameter that will take only the files older than X days (in the example 50 days)
Using that multiple times, decreasing on every execution the day range, I was able to remove all the unnecessary files
You can create a temp folder, move all the files and sub-folders you want to keep into the temp folder then delete the old folder and rename the temp folder to the old folder try this example until you are confident to do it live:
mkdir testit
cd testit
mkdir big_folder tmp_folder
touch big_folder/file1.pdf
touch big_folder/file2.pdf
mv big_folder/file1,pdf tmp_folder/
rm -r big_folder
mv tmp_folder big_folder
the rm -r big_folder will remove all files in the big_folder no matter how many. You just have to be super careful you first have all the files/folders you want to keep, in this case it was file1.pdf
To delete all *.pdf in a directory /path/to/dir_with_pdf_files/
mkdir empty_dir # Create temp empty dir
rsync -avh --delete --include '*.pdf' empty_dir/ /path/to/dir_with_pdf_files/
To delete specific files via rsync using wildcard is probably the fastest solution in case you've millions of files. And it will take care of error you're getting.
(Optional Step): DRY RUN. To check what will be deleted without deleting. `
rsync -avhn --delete --include '*.pdf' empty_dir/ /path/to/dir_with_pdf_files/
.
.
.
Click rsync tips and tricks for more rsync hacks
I found that for extremely large lists of files (>1e6), these answers were too slow. Here is a solution using parallel processing in python. I know, I know, this isn't linux... but nothing else here worked.
(This saved me hours)
# delete files
import os as os
import glob
import multiprocessing as mp
directory = r'your/directory'
os.chdir(directory)
files_names = [i for i in glob.glob('*.{}'.format('pdf'))]
# report errors from pool
def callback_error(result):
print('error', result)
# delete file using system command
def delete_files(file_name):
os.system('rm -rf ' + file_name)
pool = mp.Pool(12)
# or use pool = mp.Pool(mp.cpu_count())
if __name__ == '__main__':
for file_name in files_names:
print(file_name)
pool.apply_async(delete_files,[file_name], error_callback=callback_error)
If you want to remove both files and directories, you can use something like:
echo /path/* | xargs rm -rf
I only know a way around this.
The idea is to export that list of pdf files you have into a file. Then split that file into several parts. Then remove pdf files listed in each part.
ls | grep .pdf > list.txt
wc -l list.txt
wc -l is to count how many line the list.txt contains. When you have the idea of how long it is, you can decide to split it in half, forth or something. Using split -l command
For example, split it in 600 lines each.
split -l 600 list.txt
this will create a few file named xaa,xab,xac and so on depends on how you split it.
Now to "import" each list in those file into command rm, use this:
rm $(<xaa)
rm $(<xab)
rm $(<xac)
Sorry for my bad english.
I ran into this problem a few times. Many of the solutions will run the rm command for each individual file that needs to be deleted. This is very inefficient:
find . -name "*.pdf" -print0 | xargs -0 rm -rf
I ended up writing a python script to delete the files based on the first 4 characters in the file-name:
import os
filedir = '/tmp/' #The directory you wish to run rm on
filelist = (os.listdir(filedir)) #gets listing of all files in the specified dir
newlist = [] #Makes a blank list named newlist
for i in filelist:
if str((i)[:4]) not in newlist: #This makes sure that the elements are unique for newlist
newlist.append((i)[:4]) #This takes only the first 4 charcters of the folder/filename and appends it to newlist
for i in newlist:
if 'tmp' in i: #If statment to look for tmp in the filename/dirname
print ('Running command rm -rf '+str(filedir)+str(i)+'* : File Count: '+str(len(os.listdir(filedir)))) #Prints the command to be run and a total file count
os.system('rm -rf '+str(filedir)+str(i)+'*') #Actual shell command
print ('DONE')
This worked very well for me. I was able to clear out over 2 million temp files in a folder in about 15 minutes. I commented the tar out of the little bit of code so anyone with minimal to no python knowledge can manipulate this code.
I have faced a similar problem when there were millions of useless log files created by an application which filled up all inodes. I resorted to "locate", got all the files "located"d into a text file and then removed them one by one. Took a while but did the job!
I solved with for
I am on macOS with zsh
I moved thousands only jpg files. Within mv in one line command.
Be sure there are no spaces or special characters in the name of the files you are trying to move
for i in $(find ~/old -type f -name "*.jpg"); do mv $i ~/new; done
A bit safer version than using xargs, also not recursive:
ls -p | grep -v '/$' | grep '\.pdf$' | while read file; do rm "$file"; done
Filtering our directories here is a bit unnecessary as 'rm' won't delete it anyway, and it can be removed for simplicity, but why run something that will definitely return error?
Using GNU parallel (sudo apt install parallel) is super easy
It runs the commands multithreaded where '{}' is the argument passed
E.g.
ls /tmp/myfiles* | parallel 'rm {}'
For remove first 100 files:
rm -rf 'ls | head -100'

shell clean up script advice

Can you let me know your thoughts on this script and if you think it can be improved by any method?
I'm trying to create a clean up script that will run once a week by a cron job by root on our linux servers.
At one part of the script I call a text file that will have a list of user's names that can be deleted from, the contains of this file might change week to week.
#!/bin/bash
DAY=$(date +"%d%b%Y")
HOME='/home/user'
DOCS='/var/program/alpha/top/is'
SCRATCH='/var/program/beta/top/_temp/'
USER='/home/user/deleteuserdata.txt'
DELUSER=$USER
cd $SCRATCH
rm -rf _temp-*/
cd $DOCS
while read DELUSER; do
find $DOCS/"$DELUSER"_info* -name "*.pdf" -size +1000k -exec rm {} \;
done < $USER > $HOME/"$DAY"dellogs.txt
You should quote variables almost everywhere. Prefer pushd/popd over cd (easier to remember pervious path). Probably want to prefer find -delete over the spawn-some -exec rm. Add error checking (bash -e), and -x to see where it exits when it comes to that.
#!/bin/bash -ex
DELUSER="$USER" # setting this is useless because it's overriden in the while loop
pushd "$SCRATCH"
rm -Rf _temp-*/ || :
pushd "$DOCS"
while read DELUSER; do
find "$DOCS/$DELUSER"_info* -name "*.pdf" -size +1000k -print -delete
done <"$USER" >"$HOME/${DAY}dellogs.txt"
popd
popd

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