Extract part of file name

himu3118 · 03-26-2010, 05:46 AM

I have this string ./DAT000728-652523058.job.
I want to extract the no between DAT and - sign. I want 728. I dont want 000728.
as with
echo ./DAT000725-560162365.job | cut -d'T' -f2 | cut -d'-' -f1 I am getting 000728.

string can be ./DAT326822-652523058.job also. then i need 326822

How can I do that?

PMP · 03-26-2010, 05:54 AM

Code:

echo "DAT000728-652523058.job" | sed 's/^[^[:digit:]]\+\(.*\)-.*/\1/'

himu3118 · 03-26-2010, 06:21 AM

Thanx for the reply.

but its again giving 000728 but not 728

catkin · 03-26-2010, 06:34 AM

This task can be done very fast (if not very elegantly) using Shell Parameter Expansion with the shell configured for "extended globbing"

Code:

c@CW8:~$ string='./DAT000728-652523058.job'
c@CW8:~$ shopt -s extglob
c@CW8:~$ string=${string##*([^0-9])}
c@CW8:~$ echo $string
000728-652523058.job
c@CW8:~$ string=${string%%-*}
c@CW8:~$ echo $string
000728
c@CW8:~$ string=${string##*(0)}
c@CW8:~$ echo $string
728

Alternatively, if the character positions are fixed and only 3 digits are required

Code:

c@CW8:~$ string='./DAT000728-652523058.job'
c@CW8:~$ number=${string:8:3}
c@CW8:~$ echo $number

MTK358 · 03-26-2010, 07:27 AM

sed 's:.*DAT0*\([0-9]*\)-.*:\1:'

Code:

$ echo './DAT000728-6744645' | sed 's:.*DAT0*\([0-9]*\)-.*:\1:'
728
$ echo './DAT045728-6744645' | sed 's:.*DAT0*\([0-9]*\)-.*:\1:'
45728

himu3118 · 03-26-2010, 07:46 AM

thanx all for ur replies.

all the solutions above are working well with my script.

grail · 03-26-2010, 08:52 AM

Is this not a repeat of this previous post??

http://www.linuxquestions.org/questi...script-797130/