I have some BackUp-Files from machines which store their Backup in different folders. Additionally the files are not created at the same time (Machine 1: every sunday, Machine 2: every first monday of the month, etc.).
I need to keep the latest 10 files in each folder and delete all the others. Because of the different backup-intervals I can't just delete all files older than x days.
The folder-structure is like this:
./<SystemType>/<FQDN_Machine1>/backup_2015_09_08_02_00_00.zip
./<SystemType>/<FQDN_Machine2>/backup_2015_09_01_14_00_00.zip
IFS='
'
for i in dir/*; do
ls -d1t $i/* | head -n-10
done | xargs rm
List all subdirs excluding latest ten and send it by xargs to rm.
This is my solution:
#!/bin/bash
cat find ./ -type f | while IFS= read -r line
do
find "$line" -type f | head -n -10 | while read file
do
rm -f "$file"
done
done
Related
I need help with file handling in on Raspbian Stretch Lite running on Raspberry Pi Zero --- fresh install, updated.
The following script is run periodically as a cron job:
partition=/dev/root
imagedir=/etc/opt/kerberosio/capture/
if [[ $(df -h | grep $partition | head -1 | awk -F' ' '{ print $5/1 }' | tr ['%'] ["0"]) -gt 90 ]];
then
echo "Cleaning disk"
find $imagedir -type f | sort | head -n 100 | xargs -r rm -rf;
fi;
Essentially when the SD card is >90% full the oldest 100 files in a directory are deleted.
I want to add some functionality:
1) Copy the 100 oldest files to a NAS drive mounted on the file system and
2) Verify successful copy and
3) Delete the files that were copied.
I have found the following string which may be helpful in the modification of the script above:
find /data/machinery/capture/ -type f -name '*.*' -mtime +1 -exec mv {} /data/machinery/nas/ \;
In my shell script, I am creating a backup of my folder. I am setting this activity by cronjob and the schedule keeps on changing.
It is keeping the backup with timestamp. Like for e.g :
cd /tmp/BACKUP_DIR
backup_06-05-2014.tar
backup_06-08-2014.tar
backup_06-10-2014.tar
What I want, whenever I run the script, it should keep the latest one and the previously taken backup only. And delete the remaining backups.
Like if I run the script now, it should keep
backup_06-10-2014.tar
backup_06-18-2014.tar
And delete all the other one. What rm command should I use ?
Try as follows:
rm $(ls -1t /tmp/BACKUP_DIR | tail -n +2)
Listing sorted by date names of files with remaining only two newest
You could try deleting files older that 7 days using a find command, for example :
find /tmp/BACKUP_DIR -maxdepth 1 -type f -name "backup_*.tar" -mtime +6 -exec rm -f {} \;
Use
rm -rf `ls -lth backup_*.tar | awk '{print $NF}' | tail -n +4`
ls -lth backup_*.tar will give the sorted list of backup files (newest being at top)
awk '{print $NF}' will print file names and pass it to tail
tail -n +4 , will print file from number 3
At last tail's result is fed to rm to act
Another simplified method
rm -rf `ls -1rt backup_*.tar | tail -n +3`
I need to write a recursive script to delete all folders in a subfolder named 'date-2012-01-01_12_30' but leave the two latest.
/var/www/temp/updates/ then hundreds of folders by 'date' and by 'code'
e.g.
/var/www/temp/updates/2012-01-01/temp1/date-2012-01-_12_30
/var/www/temp/updates/2012-01-01/temp1/date-2012-02-_13_30
/var/www/temp/updates/2012-01-01/temp1/date-2013-11-_12_30
/var/www/temp/updates/2012-01-01/temp2/date-2012-01-_12_30
I was thinking about using a find to get the folder but unsure how to know what folders I can delete as the script will have to know how date- folders are in that subfolder and which ones are the latest ones
Hmm, any help would be great?
This should work:
find /var/www/temp/updates/ -type d -name "date-*" -printf '%T# %p\n' | sort -n | head -n -2 | cut -f2- | xargs rm -rf
find prints out the directory paths along with their last modification times. This is then sorted and all but the last two are deleted.
If all folders are in subdirectories temp1, temp2, ..., you can just use ls -tr
ls -dtr /var/www/temp/updates/2012-01-01/temp*/* | head -n -2 | xargs rm -rf
This lists all folders sorted by time ls -dtr, takes all but the two latest head and removes the remaining folders xargs rm -rf.
It's hard to explain in the title.
I have a bash script that runs daily to backup one folder into a zip file. The zip files are named worldYYYYMMDD.zip with YYYYMMDD being the date of backup. What I want to do is delete all but the 5 most recent backups. Sorting the files alphabetically will list the oldest ones first, so I basically need to delete all but the last 5 files when sorted in alphabetical order.
The following line should do the trick.
ls -F world*.zip | head -n -5 | xargs -r rm
ls -F: List the files alphabetically
head -n -5: Filter out all lines except the last 5
xargs -r rm: remove each given file. -r: don't run rm if the input is empty
How about this:
find /your/directory -name 'world*.zip' -mtime +5 | xargs rm
Test it before. This should remove all world*.zip files older than 5 days. So a different logic than you have.
I can't test it right now because I don't have a Linux machine, but I think it should be:
rm `ls -A | head -5`
ls | grep ".*[\.]zip" | sort | tail -n-5 | while read file; do rm $file; done
sort sorts the files
tail -n-5 returns all but the 5 most recent
the while loop does the deleting
ls world*.zip | sort -r | tail n+5 | xargs rm
sort -r will sort in reversed order, so the newest will be at the top
tail n+5 will output lines, starting with the 5th
xargs rm will remove the files. Xargs is used to pass stdin as parameters to rm.
I'm using bash.
Suppose I have a log file directory /var/myprogram/logs/.
Under this directory I have many sub-directories and sub-sub-directories that include different types of log files from my program.
I'd like to find the three newest files (modified most recently), whose name starts with 2010, under /var/myprogram/logs/, regardless of sub-directory and copy them to my home directory.
Here's what I would do manually
1. Go through each directory and do ls -lt 2010*
to see which files starting with 2010 are modified most recently.
2. Once I go through all directories, I'd know which three files are the newest. So I copy them manually to my home directory.
This is pretty tedious, so I wondered if maybe I could somehow pipe some commands together to do this in one step, preferably without using shell scripts?
I've been looking into find, ls, head, and awk that I might be able to use but haven't figured the right way to glue them together.
Let me know if I need to clarify. Thanks.
Here's how you can do it:
find -type f -name '2010*' -printf "%C#\t%P\n" |sort -r -k1,1 |head -3 |cut -f 2-
This outputs a list of files prefixed by their last change time, sorts them based on that value, takes the top 3 and removes the timestamp.
Your answers feel very complicated, how about
for FILE in find . -type d; do ls -t -1 -F $FILE | grep -v "/" | head -n3 | xargs -I{} mv {} ..; done;
or laid out nicely
for FILE in `find . -type d`;
do
ls -t -1 -F $FILE | grep -v "/" | grep "^2010" | head -n3 | xargs -I{} mv {} ~;
done;
My "shortest" answer after quickly hacking it up.
for file in $(find . -iname *.php -mtime 1 | xargs ls -l | awk '{ print $6" "$7" "$8" "$9 }' | sort | sed -n '1,3p' | awk '{ print $4 }'); do cp $file ../; done
The main command stored in $() does the following:
Find all files recursively in current directory matching (case insensitive) the name *.php and having been modified in the last 24 hours.
Pipe to ls -l, required to be able to sort by modification date, so we can have the first three
Extract the modification date and file name/path with awk
Sort these files based on datetime
With sed print only the first 3 files
With awk print only their name/path
Used in a for loop and as action copy them to the desired location.
Or use #Hasturkun's variant, which popped as a response while I was editing this post :)