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UNIX scripting - ksh - get part of filename as valiable

Hi,

Would like to see somebody's help on writing small ksh script runs on UNIX.

Environment : Solaris 8/10
in one directory say /var/tmp , there are multiple files in below manner.
abc_<date>
i.e. abc_20110501 , abc_20110502, abc_20110503

I would like to get "latest file's filename , only date part, for this case,   only for "20110503"
within ksh script.

Would somebody kindly help?
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sleepingluke
Asked:
sleepingluke
2 Solutions
 
TintinCommented:
cd /var/tmp
file=$(ls abc*)
echo ${file#*_}

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TintinCommented:
Small correction

file=$(ls abc*)

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should be

file=$(ls -t abc*)

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assuming you're basing the latest file on modification date.
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theKashyapCommented:
Tintin's soln assumes that the timestamps on the file are matching the timestamps in filename. Without that it won't work. E.g. if you transfer the files to a different place.

Use this if you really want to sort by the timestamp given in the file name:
// assuming YYYYMMDD format:
LATEST_DATE=`ls -1 /var/tmp/abc_* | sed 's/abc_//' | sort | tail -1`
// or you can use cut as well if the prefix is fixed length (always abc_)
LATEST_DATE=`ls -1 /var/tmp/abc_* | cut -b 5- | sort | tail -1`

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Gerwin Jansen, EE MVETopic Advisor Commented:
LATEST_DATE=`ls -1tr abc_* | tail -1 | cut -d"_" -f2`

explanation:

ls -1tr abc_*
# list all abc_* files in 1 column, sorted by date, most recent file last

tail -1
# get only last file = most recent file

cut -d"_" -f2
# cut string at _ and print 2nd part (date
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sleepinglukeAuthor Commented:
thanks all for helpful tips and knowledges!
I could workout with theKashyap second command as date should be fixed length for a time being.

Appreciated!
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tel2Commented:
Note that you don't need ls's "-1" switch in this context (i.e. just "ls" will give the same result here).
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