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find command: various date ranges

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# 1  
Old 10-24-2008
find command: various date ranges
Hi,

I have writtena script that will recursivly go into subdirecotries and report out what files there are in there that have not been accessed over various date ranges.

I do this using a number of find commands:
Code:
find . -path './.snapshot' -prune -o  -type f             -atime -8   
find . -path './.snapshot' -prune -o  -type f -atime +7   -atime -32  
find . -path './.snapshot' -prune -o  -type f -atime +31  -atime -187 
find . -path './.snapshot' -prune -o  -type f -atime +186 -atime -366 
find . -path './.snapshot' -prune -o  -type f -atime +365

I know this is highly inefficient - and it takes ages! Any suggestions on what I can do to speed this up?
As you can see I am having to ezxclude snapshots and am only interested in file sizes. For inforation, I then pipe this through an awk command to sum the total bytesand number of files.
# 2  
Old 10-24-2008
Are you on Linux?
# 3  
Old 10-24-2008
Ideally I'd be running this script on both Linux and Unix machines, but realise that this is going to make it harder, so trying to stick to Linux for now.

So in short - yes.
# 4  
Old 10-24-2008
I am not on Linux now - so I cannot give you a working answer in code form.
pseudo code:
use the stat command to produce the filename and the atime in epoch seconds
I think this is close to correct for the command I want - you can check it:
Code:
stat -c "%N %x" -t "%s" filename

This is what I need Linux to verify.
Code:
now=date "+%s"
find -path './snapshot' -prune -o -type f -exec stat -c "%x %N" -t "%s" {} | \
while read asecs filename
do
     daysago=$(( (now - asecs) / 86400  ))
     monthsago=$(( daysago / 30 ))
     echo "$daysago $monthsago $filename"
done | sort -n > outputfile

You now have a file you can work with using awk - atime in days before today, atime in months before today, and file name. The reason you do this is because you want to find all files just once. Repeating find over and over agsain on the same filesystem is a waste of time. And is why your code is slow.

Example: distribution of file ages in months order by months ago
Code:
awk ' { arr[$2]++ } END { for (i in arr) { print i, arr[i]} ' outputfile | sort -n

# 5  
Old 10-24-2008
Thanks. no time to check it right now - but will try on Monday.
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