I have RaspBerry Pi project where A/D converter collects 5 minutes long data files into data folder and all data except newest one will be removed in every hour by dataFlush.py script and crontab. The path of dataFlush.py is "/home/pi" and the path of the data folder is "/home/pi/dataLog". So how do I add dataLog path into this command:
os.system('ls -t | tail -n +2 | xargs rm --')
I tried something like this but it didn't work:
os.system('ls /home/pi/dataLog -t | tail -n +2 | xargs rm --')
ls(1) won't generate full-path outputs in your case, so you might want to prepend every line with the "prefix":
os.system('ls /home/pi/dataLog -t | tail -n +2 | sed 's|^|/home/pi/dataLog/|' | xargs rm --')
Another options is to use find(1):
os.system('find /home/pi/dataLog | xargs ls -t -- | tail -n +2 | xargs rm --')
How would I do the following command? This is the following pseudocode I have:
ls -U | head -4 < rm *
I want to delete the below four files
ubuntu#ip:/tmp$ ls -U | head -4
c9e5aff2-f4f6-459e-9752-466625165fa0.jpg
815626f8-2650-4596-94b9-796a82780390.jpg
b9259a6b-b421-4b64-ba4d-72fd632703a3.jpg
78c243e7-43df-417d-9dcf-a4578f1c1f20.jpg
xargs should do the trick:
$ ls -U | head -4 | xargs rm
If you can't remember bash syntax a simple for loop often works like a charm:
for file in `ls -U | head -4`; do rm $file; done
This is a backup script that copies files from one directory to another. I use a for loop to check if there are more than five files. If there are, the loop should delete the oldest entries first.
I tried ls -tr | head -n -5 | xargs rm from the command line and it works successfully to delete older files if there are more than 5 in the directory.
However, when I put it into my for loop, I get an error rm: missing operand
Here is the full script. I don't think I am using the for loop correctly in the script, but I'm really not sure how to use the commands ls -tr | head -n -5 | xargs rm in a loop that iterates over the files in the directory.
timestamp=$(date +"%m-%d-%Y")
dest=${HOME}/mybackups
src=${HOME}/safe
fname='bu_'
ffname=${HOME}/mybackups/${fname}${timestamp}.tar.gz
# for loop for deletion of file
for f in ${HOME}/mybackups/*
do
ls -tr | head -n -5 | xargs rm
done
if [ -e $ffname ];
then
echo "The backup for ${timestamp} has failed." | tee ${HOME}/mybackups/Error_${timestamp}
else
tar -vczf ${dest}/${fname}${timestamp}.tar.gz ${src}
fi
Edit: I took out the for loop, so it's now just:
[...]
ffname=${HOME}/mybackups/${fname}${timestamp}.tar.gz
ls -tr | head -n -5 | xargs rm
if [ -e $ffname ];
[...]
The script WILL work if it is in the mybackups directory, however, I continue to get the same error if it is not in that directory. The script gets the file names but tries to remove them from the current directory, I think... I tried several modifications but nothing has worked so far.
I get an error rm: missing operand
The cause of that error is that there are no files left to be deleted. To avoid that error, use the --no-run-if-empty option:
ls -tr | head -n -5 | xargs --no-run-if-empty rm
In the comments, mklement0 notes that this issue is peculiar to GNU xargs. BSD xargs will not run with an empty argument. Consequently, it does not need and does not support the --no-run-if-empty option.
More
Quoting from a section of code in the question:
for f in ${HOME}/mybackups/*
do
ls -tr | head -n -5 | xargs rm
done
Note that (1) f is never used for anything and (2) this runs the ls -tr | head -n -5 | xargs rm several times in a row when it needs to be run only once.
Obligatory Warning
Your approach parses the output of ls. This makes for a simple and easily understood command. It can work if all your files are sensibly named. It will not work in general. For more on this, see: Why you shouldn't parse the output of ls(1).
Safer Alternative
The following will work with all manner of file names, whether they contains spaces, tabs, newlines, or whatever:
find . -maxdepth 1 -type f -printf '%T# %i\n' | sort -n | head -n -5 | while read tstamp inode
do
find . -inum "$inode" -delete
done
SMH. I ended up coming up to the simplest solution in the world by just cd-ing into the directory before I ran ls -tr | head -n -5 | xargs rm . Thanks for everyone's help!
timestamp=$(date +"%m-%d-%Y")
dest=${HOME}/mybackups
src=${HOME}/safe
fname='bu_'
ffname=${HOME}/mybackups/${fname}${timestamp}.tar.gz
cd ${HOME}/mybackups
ls -tr | head -n -5 | xargs rm
if [ -e $ffname ];
then
echo "The backup for ${timestamp} has failed." | tee ${HOME}/mybackups/Error_${timestamp}
else
tar -vczf ${dest}/${fname}${timestamp}.tar.gz ${src}
fi
This line ls -tr | head -n -5 | xargs rm came from here
ls -tr displays all the files, oldest first (-t newest first, -r
reverse).
head -n -5 displays all but the 5 last lines (ie the 5 newest files).
xargs rm calls rm for each selected file
.
i have a lot of different type of files in one folder. i need to delete the files but except the pdf file.
I tried to display the pdf file only. but i need to delete the other than pdf files
ls -1 | xargs file | grep 'PDF document,' | sed 's/:.*//'
You could do the following - I've used echo rm instead of rm for safety:
for i in *
do
[ x"$(file --mime-type -b "$i")" != xapplication/pdf ] && echo rm "$i"
done
The --mime-type -b options to file make the output of file easier to deal with in a script.
$ ls
aa.txt a.pdf bb.cpp b.pdf
$ ls | grep -v .pdf | xargs rm -rf
$ ls
a.pdf b.pdf
:) !
ls |xargs file|awk -F":" '!($2~/PDF document/){print $1}'|xargs rm -rf
Try inverting the grep match:
ls -1 | xargs file | grep -v 'PDF document,' | sed 's/:.*//'
It's rare in my experience to encounter PDF files which don't have a .pdf extension. You don't state why "file" is necessary in the example, but I'd write this as:
# find . -not -name '*.pdf' -delete
Note that this will recurse into subdirectories; use "-maxdepth 1" to limit to the current directory only.
I am new to shell scripting, so I need some help here. I have a directory that fills up with backups. If I have more than 10 backup files, I would like to remove the oldest files, so that the 10 newest backup files are the only ones that are left.
So far, I know how to count the files, which seems easy enough, but how do I then remove the oldest files, if the count is over 10?
if [ls /backups | wc -l > 10]
then
echo "More than 10"
fi
Try this:
ls -t | sed -e '1,10d' | xargs -d '\n' rm
This should handle all characters (except newlines) in a file name.
What's going on here?
ls -t lists all files in the current directory in decreasing order of modification time. Ie, the most recently modified files are first, one file name per line.
sed -e '1,10d' deletes the first 10 lines, ie, the 10 newest files. I use this instead of tail because I can never remember whether I need tail -n +10 or tail -n +11.
xargs -d '\n' rm collects each input line (without the terminating newline) and passes each line as an argument to rm.
As with anything of this sort, please experiment in a safe place.
find is the common tool for this kind of task :
find ./my_dir -mtime +10 -type f -delete
EXPLANATIONS
./my_dir your directory (replace with your own)
-mtime +10 older than 10 days
-type f only files
-delete no surprise. Remove it to test your find filter before executing the whole command
And take care that ./my_dir exists to avoid bad surprises !
Make sure your pwd is the correct directory to delete the files then(assuming only regular characters in the filename):
ls -A1t | tail -n +11 | xargs rm
keeps the newest 10 files. I use this with camera program 'motion' to keep the most recent frame grab files. Thanks to all proceeding answers because you showed me how to do it.
The proper way to do this type of thing is with logrotate.
I like the answers from #Dennis Williamson and #Dale Hagglund. (+1 to each)
Here's another way to do it using find (with the -newer test) that is similar to what you started with.
This was done in bash on cygwin...
if [[ $(ls /backups | wc -l) > 10 ]]
then
find /backups ! -newer $(ls -t | sed '11!d') -exec rm {} \;
fi
Straightforward file counter:
max=12
n=0
ls -1t *.dat |
while read file; do
n=$((n+1))
if [[ $n -gt $max ]]; then
rm -f "$file"
fi
done
I just found this topic and the solution from mikecolley helped me in a first step. As I needed a solution for a single line homematic (raspberrymatic) script, I ran into a problem that this command only gave me the fileames and not the whole path which is needed for "rm". My used CUxD Exec command can not start in a selected folder.
So here is my solution:
ls -A1t $(find /media/usb0/backup/ -type f -name homematic-raspi*.sbk) | tail -n +11 | xargs rm
Explaining:
find /media/usb0/backup/ -type f -name homematic-raspi*.sbk searching only files -type f whiche are named like -name homematic-raspi*.sbk (case sensitive) or use -iname (case insensitive) in folder /media/usb0/backup/
ls -A1t $(...) list the files given by find without files starting with "." or ".." -A sorted by mtime -t and with a return of only one column -1
tail -n +11 return of only the last 10 -n +11 lines for following rm
xargs rm and finally remove the raiming files in the list
Maybe this helps others from longer searching and makes the solution more flexible.
stat -c "%Y %n" * | sort -rn | head -n +10 | \
cut -d ' ' -f 1 --complement | xargs -d '\n' rm
Breakdown: Get last-modified times for each file (in the format "time filename"), sort them from oldest to newest, keep all but the last ten entries, and then keep all but the first field (keep only the filename portion).
Edit: Using cut instead of awk since the latter is not always available
Edit 2: Now handles filenames with spaces
On a very limited chroot environment, we had only a couple of programs available to achieve what was initially asked. We solved it that way:
MIN_FILES=5
FILE_COUNT=$(ls -l | grep -c ^d )
if [ $MIN_FILES -lt $FILE_COUNT ]; then
while [ $MIN_FILES -lt $FILE_COUNT ]; do
FILE_COUNT=$[$FILE_COUNT-1]
FILE_TO_DEL=$(ls -t | tail -n1)
# be careful with this one
rm -rf "$FILE_TO_DEL"
done
fi
Explanation:
FILE_COUNT=$(ls -l | grep -c ^d ) counts all files in the current folder. Instead of grep we could use also wc -l but wc was not installed on that host.
FILE_COUNT=$[$FILE_COUNT-1] update the current $FILE_COUNT
FILE_TO_DEL=$(ls -t | tail -n1) Save the oldest file name in the $FILE_TO_DEL variable. tail -n1 returns the last element in the list.
Based on others suggestions and some awk foo, I got this to work. I know this an old thread, but I didn't find a decent answer here and this sorted it for me. This just deletes the oldest file, but you can change the head -n 1 to 10 and get the oldest 10.
find $DIR -type f -printf '%T+ %p\n' | sort | head -n 1 | awk '{first =$1; $1 =""; print $0}' | xargs -d '\n' rm
Using inode numbers via stat & find command (to avoid pesky-chars-in-file-name issues):
stat -f "%m %i" * | sort -rn -k 1,1 | tail -n +11 | cut -d " " -f 2 | \
xargs -n 1 -I '{}' find "$(pwd)" -type f -inum '{}' -print
#stat -f "%m %i" * | sort -rn -k 1,1 | tail -n +11 | cut -d " " -f 2 | \
# xargs -n 1 -I '{}' find "$(pwd)" -type f -inum '{}' -delete