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

Shell Programming and Scripting
1

there are mutiple file nams in the directory. How to return the the lastest files for each file name.

ex.

abc1234_050201
abc1234_050206
abc1234_050208
xyz34_050204
xyz34_050210
xyz34_050218

thanks

2

Try:

ls | sort -t_ -k2,2rn | awk -F_ '!A[$1]++'
3

the script still rturns lots file name but not the one for each name pattern. thanks

4

This works for me:

 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

With your sample file names I get:

ls | sort -t_ -k2,2rn | awk -F_ '!A[$1]++'
xyz34_050218
abc1234_050208

What do you get?

Otherwise try:

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

or

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

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