bash script to remove previous backup - bash

Question

I set up a daily cron job to backup my server.
In my folder backup, the backup command generates 2 files : the archive itself .tar.gz and a file .info.json like the ones below:
-rw-r--r-- 1 root root 1617 Feb 2 16:17 20200202-161647.info.json
-rw-r--r-- 1 root root 48699726 Feb 2 16:17 20200202-161647.tar.gz
-rw-r--r-- 1 root root 1617 Feb 3 06:25 20200203-062501.info.json
-rw-r--r-- 1 root root 48737781 Feb 3 06:25 20200203-062501.tar.gz
-rw-r--r-- 1 root root 1618 Feb 4 06:25 20200204-062501.info.json
-rw-r--r-- 1 root root 48939569 Feb 4 06:25 20200204-062501.tar.gz
How to I write a bash script that will only keep the last 2 archives and deletes all the others backup (targ.gz and info.json).
In this example, that would mean deleted 20200204-062501.info.json and 20200204-062501.tar.gz .
Edit:
I replace -name by -wholename in the script but when I run it, it doesn't have any effects apparently.The old archives are still there and they have not been deleted.
the script :
#!/bin/bash
DEBUG="";
DEBUG="echo DEBUG..."; #put last to safely debug without deleting files
keep=2;
for suffix in /home/archives .json .tar; do
list=( $( find . -wholename "*$suffix" ) ); #allow for zero names
if [ ${#list[#]} -gt $keep ]; then
# delete all but last $keep oldest files
${DEBUG}rm -f "$( ls -tr "${list[#]}" | head -n-$keep )";
fi
done
Edit 2:
if I run #sorin script, does it actually delete everything if I believe the script output?
The archive folder before running the script:
https://pastebin.com/7WtwVHCK
The script I run:
find home/archives/ \( -name '*.json' -o -name '*.tar.gz' \) -print0 |\
sort -zr |\
sed -z '3,$p' | \
xargs -0 echo rm -f
The script output:
https://pastebin.com/zd7a2zcq
Edit 3 :
The command find /home/archives/ -daystart \( -name '*.json' -o -name '*.tar.gz' \) -mtime +1 -exec echo rm -f {} + works and does the job.
Marked as solved

Answer 1

If the file is generated daily, a simple approach would be to take advantage of the -mtime find condition:
find /home/archives/ -daystart \( -name '*.json' -o -name '*.tar.gz' \) -mtime +1 -exec echo rm -f {} +
-daystart - use the start of the day for comparing modification times
\( -name '*.json' -o -name '*.tar.gz' \) - select files that end either in *.json or *.tar.gz
-mtime +1 - modification time is older than 24 hours (from the day start)
-exec echo rm -f {} + - remove the files (remove the echo after testing and verifying the result is what you want)
A simpler solution avoiding ls and it's pitfalls and not depending on the modification time of the files:
find /home/archives/ \( -name '*.json' -o -name '*.tar.gz' \) -print0 |\
sort -zr |\
sed -nz '3,$p' | \
xargs -0 echo rm -f
\( -name '*.json' -o -name '*.tar.gz' \) - find files that end in either *.json or tar.gz
-print0 - print them null separated
sort -zr - -z tells sort to use null as a line separator, -r sorts them in reverse
sed -nz '3,$p' - -z same as above. '3,$p' - print lines between 3rd and the end ($)
xargs -0 echo rm -f - execute rm with the piped arguments (remove the echo after you tested and you are satisfied with the command)
Note: not all sort and sed support the -z but most do. If you are stuck with such a situation, you might have to use a higher level language

Answer 2

Find the two most recent files in path:
most_recent_json=$(ls -t *.json | head -1)
most_recent_tar_gz=$(ls -t *.tar.gz | head -1)
Remove everything else ignoring the found recent files:
rm -i $(ls -I $most_recent_json -I $most_recent_tar_gz)

Answer 3

Automatic deleting can be hazardous to your mental state if it deletes unwanted files or aborts long scripts early due to unexpected errors. Say when there are fewer than 1+2 files in your example. Be sure the script does not fail if there are no files at all.
tdir=/home/archives/; #target dir
DEBUG="";
DEBUG="echo DEBUG..."; #put last to safely debug without deleting files
keep=2;
for suffix in .json .tar; do
list=( $( find "$tdir" -name "*$suffix" ) ); #allow for zero names
if [ ${#list[#]} -gt $keep ]; then
# delete all but last $keep oldest files
${DEBUG}rm -f "$( ls -tr "${list[#]}" | head -n-$keep )";
fi
done

Answer 4

Assuming that you have fewer than 10 files and that they are created in pairs, then you can do something straightforward like this:
files_to_delete=$(ls -tr1 | tail -n+3)
rm $files_to_delete
The -tr1 tells the ls command to list the files in reverse chronological order by modification time, each on a single line.
The tail -n+3 tells the tail command to start at the third line (skipping the first two lines).
If you have more than 10 files, a more complicated solution will be necessary, or you would need to run this multiple times.