Re: HowTo : Extract word with 'sed'
On Fri, 9 Aug 2002, Paul Branston wrote:
> On Thu, Aug 08, 2002 at 11:40:32AM -0400, Michael C Tiernan wrote:
> > On Thursday 08 August 2002 08:59, Nick Lindsell said:
> > > At 14:34 08/08/2002 +0200, you wrote:
> > > > "File_21_05082002" and i would like to
> > > > extract "21" from this.
> > > >How can i do it with 'sed'?
> > >
> > > No need for sed, cut is simpler:-
> > > $extract= echo "File_21_05082002"|cut -c 6-7
> >
> > Cut is easier for this problem but I'd modify it one bit
> > $extract= `echo "File_21_05082002"|cut -d_ -f2`
> >
> > This assumes (*cough*) that you want the value between the two "_" characters.
> >
> > To answer your original question. In sed:
> > $extract=`echo $Fname | sed -e "s/^File_//" -e "s/_[0-9]*$//"`
> >
> > Should come very close to what you wanted (I think...)
>
> or in sed using 1 pattern match assuming File_ always comes before the
> digits and the end of the digits is a _ character.
>
> echo "File_21_05082002"|sed -e 's/^File_\(.*\)_.*$/\1/'
>
Or just use shell globbing:
orig="File_21_05082002"
temp=${orig#*_}
result=${temp%_*}
echo $result
--
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Re: HowTo : Extract word with 'sed'
On Thu, Aug 08, 2002 at 11:40:32AM -0400, Michael C Tiernan wrote: > On Thursday 08 August 2002 08:59, Nick Lindsell said: > > At 14:34 08/08/2002 +0200, you wrote: > > > "File_21_05082002" and i would like to > > > extract "21" from this. > > >How can i do it with 'sed'? > > > > No need for sed, cut is simpler:- > > $extract= echo "File_21_05082002"|cut -c 6-7 > > Cut is easier for this problem but I'd modify it one bit > $extract= `echo "File_21_05082002"|cut -d_ -f2` > > This assumes (*cough*) that you want the value between the two "_" characters. > > To answer your original question. In sed: > $extract=`echo $Fname | sed -e "s/^File_//" -e "s/_[0-9]*$//"` > > Should come very close to what you wanted (I think...) or in sed using 1 pattern match assuming File_ always comes before the digits and the end of the digits is a _ character. echo "File_21_05082002"|sed -e 's/^File_\(.*\)_.*$/\1/' -- redhat-list mailing list unsubscribe mailto:[EMAIL PROTECTED]?subject=unsubscribe https://listman.redhat.com/mailman/listinfo/redhat-list
Re: HowTo : Extract word with 'sed'
On Thu, Aug 08, 2002 at 01:59:50PM +0100, Nick Lindsell wrote: > At 14:34 08/08/2002 +0200, you wrote: > > > I have a word which is :"File_21_05082002" and i would like to > > extract "21" from this. > > > >How can i do it with 'sed'? > > No need for sed, cut is simpler:- > > $extract= echo "File_21_05082002"|cut -c 6-7 cut is simpler, but if the format is always word_digits_word, then cutting based on exact character count might not be best solution. You can also cut based on a delimiter. Try (in bash): extract=`echo "File_21_05082002"|cut -d _ -f 2` This cuts the 2nd field where the string is delimited by underscores. -- Ed Wilts, Mounds View, MN, USA mailto:[EMAIL PROTECTED] Member #1, Red Hat Community Ambassador Program -- redhat-list mailing list unsubscribe mailto:[EMAIL PROTECTED]?subject=unsubscribe https://listman.redhat.com/mailman/listinfo/redhat-list
Re: HowTo : Extract word with 'sed'
On Thursday 08 August 2002 08:59, Nick Lindsell said: > At 14:34 08/08/2002 +0200, you wrote: > > "File_21_05082002" and i would like to > > extract "21" from this. > >How can i do it with 'sed'? > > No need for sed, cut is simpler:- > $extract= echo "File_21_05082002"|cut -c 6-7 Cut is easier for this problem but I'd modify it one bit $extract= `echo "File_21_05082002"|cut -d_ -f2` This assumes (*cough*) that you want the value between the two "_" characters. To answer your original question. In sed: $extract=`echo $Fname | sed -e "s/^File_//" -e "s/_[0-9]*$//"` Should come very close to what you wanted (I think...) -- redhat-list mailing list unsubscribe mailto:[EMAIL PROTECTED]?subject=unsubscribe https://listman.redhat.com/mailman/listinfo/redhat-list
