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Shell Programming and Scripting

Sorting prob

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# 1  
Old 08-04-2006
Sorting prob
Hi

I have a directory having following three set of files as follows:

test.20060804
test.20060801
test.20060802

I want to list the latest file which is less than 20060803. in this case, it will be test.20060802.

How can i achieve the same? Any pointers to above will be appreciated.

Thanks.
# 2  
Old 08-04-2006
something to start with:
Code:
ls test.* | nawk -F'.' '{print $NF, $0}' | sort -rn | cut -d ' ' -f2-

# 3  
Old 08-04-2006
thanks but...
Thanks for your reply. I understand you are trying to put the date part in first column in list output. But then how do you select the correct file? My requirement is to somehow pass '20060803' and get the file which is less than that date. If the result is more than one file, then pick the latest one. So, in this case, it will be test.20060802

Thanks for your help.
# 4  
Old 08-04-2006
Perhaps something like this:
Code:
ls test.* | 
nawk -F. -v target_dt=20060803 '
    {
        if ($2 < target_dt && max_dt < $2) {
            max_dt=$2
            target_file=$0
    }
}
END {
    print target_file
}'

Results:
Code:
test.20060802

# 5  
Old 08-04-2006
how 'bout.....
Code:
ls test.* | nawk -F'.' -v pivot='20060803' '$NF < pivot && $NF > latestDate {latest=$0;latestDate=$NF} END {print latest}'

# 6  
Old 08-04-2006
tmarikle,
what can I say...... great minds........
# 7  
Old 08-04-2006
thanks
You guys are awesome !!!
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