shell script for deleting files, where files have sequence numbers - linux

Question

I need a shell script for deleting files, where files have sequence numbers.
For example, there is a directory like /abc/def inside of which I have files like:
xyz_1000_1_pqr.arc
xyz_1001_1_pqr.arc
xyz_1002_1_pqr.arc
xyz_1003_1_pqr.arc
xyz_1004_1_pqr.arc
xyz_1005_1_pqr.arc
xyz_1006_1_pqr.arc
xyz_1007_1_pqr.arc
xyz_1008_1_pqr.arc
Here I need to delete all the .arc files that have sequence number less than(<) 1004. That is, only files:
xyz_1000_1_pqr.arc
xyz_1001_1_pqr.arc
xyz_1002_1_pqr.arc
xyz_1003_1_pqr.arc
should be deleted.
(P.S : each file is 4-5GB and critical )
EDIT:
sorry for not mentioning earlier
Requirements:deleting the files that have been backed up and preventing the file system reaching 100% ,the backup team provides the sequence number of latest file that as been backed up
it would be very convinient if i can get a shell script that takes an argument(sequence number of the last backed up file) and delete all the files having sequence number less than the sequence number provided by the back up team
there are more than 30 servers on which i have the same scenario,and the starting sequence number(sequence number of the oldest file ) will be different for each and will not be known unless logging into each and checking the directory manualy
hence a for loop with a starting sequence number till ending sequence and rm is out of the question
A generic script that can be deployed on all servers which can work only with the sequence number of the file that has been backed up recently,is what im looking for so that it can be called as a event -reaction from a tool(OEM12c oracle related that generates Filesystem alerts )
As of now I am logging int each server manually removing them using regular expressions every time a alert is triggered file system crosses 70% which is repetitive and hectic as I have other concerns of (DBA), hence a automated script would save me a lot of time .
thanks

Answer 1

One way
rm xyz_100{0..3}_1_pqr.arc
if you have the start and end sequence number, its just a matter of looping over and deleting them
for (( i=$start_num ; i<=$end_num; i++ ))
do
rm xyz_${i}_*arc
done

Answer 2

This script starts with the number that you specify on the command line and deletes all files with that number or lower until it gets down to a number where the file doesn't exist.
#!/bin/sh
num=$1
for ((i=$num; i>=0; i--))
do
name=$(printf 'xyz_%03i_1_pqr.arc' $i)
[ -f "$name" ] || break
rm "$name"
done
The "%03i" format in the printf statement assures that the number, once formatted, will be three digits or longer. (That means that a number such as 99 is padded with zeros to become 099.) For printf, "%i" would mean format an integer, "%3i" would mean format an integer and give it at least three spaces, and "%03i" means format an integer into three spaces, left-padded with zeros as needed.
In an earlier version of this answer, I had the script check all numbers down to zero looking for files to delete. In the comments, you mention that sequence numbers may be up to 7 digits. That could make the exhaustive approach overly time consuming. In this version, I have it count down until it reaches a sequence number for which the backup has already been deleted and it stops there.