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Retaining file versions based on input parameter

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# 1  
Old 07-15-2013
Retaining file versions based on input parameter
Hello Forum.

I have the following files in one directory:

Code:
abc_july01_2013.txt
abc_july02_2013.txt
abc_july03_2013.txt
abc_july04_2013.txt
abc_july05_2013.txt
abc_july06_2013.txt
abc_july07_2013.txt
abc_july08_2013.txt

If I want to be able to keep the last 5 versions of the file and delete the rest so that all that remains is:

Code:
abc_july04_2013.txt
abc_july05_2013.txt
abc_july06_2013.txt
abc_july07_2013.txt
abc_july08_2013.txt

Then the following code will accomplish this:
Code:
rm -f `ls -tp abc* | grep -v "/$" | awk '{ if (NR > 5) print; }'`

But I don't want to hard-code the # of versions (which is 5 in this case). I want to be able to pass in a parameter value to the command so that it's more flexible. Is there an easy way that it can be done?

Thanks
Last edited by Scott; 07-15-2013 at 04:51 PM.. Reason: More code tags
# 2  
Old 07-15-2013
Code:
if [ -z "$1" ] 
then
        echo "Must pass in a number" >&2
        exit 1
fi

ls -tp abc* | grep -v "/$" | -v MAX="$1" awk 'NR > MAX' | xargs echo rm

Remove the echo once you've tested and are sure it does what you want.
# 3  
Old 07-15-2013
Thank you Corona688 for your quick response.

I tried your code but I'm getting the following error:

Code:
ksh: -v:  not found.

Is that -v just before the MAX variable intentional?
# 4  
Old 07-15-2013
Whoops, typo.

Code:
ls -tp abc* | grep -v "/$" | awk -v MAX="$1" 'NR > MAX' | xargs echo rm

This User Gave Thanks to Corona688 For This Post:
pchang
# 5  
Old 07-15-2013
Thank you Fellow Canadian - it works beautifully. Smilie
This User Gave Thanks to pchang For This Post:
Corona688
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