Hello,

how do i get the last field
Ex: files are with this formats
abc_xyz_20041232
abc_20041001
abbc_xyz_rrr_20041013

i tried with cut command but i doono how to get the last filed the file names has many dilimiters "_"

any help appreciated

Thanks
Mehul

You have these filenames contained in a variable, as if within a for filename loop?

last=$(echo $filename | awk -F_ '{print $NF}')

echo $last
20041232

for filename in `cat ./tmp`
do
last=$(echo $filename | awk -F_ '{print $NF}')
done

It gives syntax error
syntax error at line 13: `last=$' unexpected

it works fine with
echo $filename | awk -F_ '{print $NF}'

but not able to store the value

Another way to do it:

#!/bin/ksh

for i in abc_xyz_20041232 abc_20041001 abbc_xyz_rrr_20041013
do
set - $(IFS="_"; echo $i)
shift $(($#-1))
var=$1
echo $var
done

Use the set command to parse the string into command line arguments. Then shift the total number of variables -1 and that sets $1, first argument, equal to the last argument.

Regards,

Nails

Mehul,

My code requires ksh. If you are not using ksh, then use backquotes instead:

last=`echo $filename | awk -F_ '{print $NF}'`

Let me know how that works.

to avoid awk just use:

{ while read line
do
LAST=`echo ${line##*_}`
done } < your_file

LAST will have the last field.

And, if you are always chopping off only the last 8 characters:

#!/bin/ksh

typeset -R8 line
exec 3<./data.file

while read -u3 line
do
print $line
done