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There are multiple filenames in the directory.How to return the the lastest files for each file name

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Top Forums Shell Programming and Scripting There are multiple filenames in the directory.How to return the the lastest files for each file name
# 1  
Old 02-18-2015
There are multiple filenames in the directory.How to return the the lastest files for each file name
there are mutiple file nams in the directory. How to return the the lastest files for each file name.

ex.

Code:
abc1234_050201
abc1234_050206
abc1234_050208
xyz34_050204
xyz34_050210
xyz34_050218

thanks
Last edited by Scrutinizer; 02-18-2015 at 04:10 PM.. Reason: code tags
# 2  
Old 02-18-2015
Try:
Code:
ls | sort -t_ -k2,2rn | awk -F_ '!A[$1]++'

# 3  
Old 02-18-2015
the script still rturns lots file name but not the one for each name pattern. thanks
# 4  
Old 02-18-2015
This works for me:

Code:
✔ tmp1 $ ls
34567_a  34567_b  34567_c  34567_d  abc_a  abc_b  abc_c  abc_d  ascript.sh

+ tmp1 $ bash ascript.sh 
34567_d
abc_d

+ tmp1 $ cat ascript.sh 
newlist=""
for f in *_*;do
	nl=${f/_*/}
	echo "$newlist" | grep -q "$nl" || newlist+=" $nl"
done

for nl in $(echo $newlist|sort -u);do
	for this in $nl*;do echo > /dev/zero;done
	echo $this
done

hth
# 5  
Old 02-18-2015
With your sample file names I get:
Code:
ls | sort -t_ -k2,2rn | awk -F_ '!A[$1]++'
xyz34_050218
abc1234_050208

What do you get?

Otherwise try:
Code:
ls xyz*_* abc*_* | sort -t_ -k2,2rn | awk -F_ '!A[$1]++'
xyz34_050218
abc1234_050208

or
Code:
ls xyz*_* abc*_* | sort -t_ -k2,2rn | awk -F_ '!A[$1]++'

Code:
\ls xyz*_* abc*_* | sort -t_ -k2,2rn | awk -F_ '!A[$1]++'

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