topic Re: Parameter substitution in shell: does ${##} work ? in Operating System - Linux

Parameter substitution in shell: does ${##} work ?

CAS_2 — Tue, 24 Oct 2006 05:57:29 GMT

Hi

Let be a numeric value into a shell variable:

N=000034

I'm interested in removing leading zeroes.
According to man pages, sh-posix can do this task by means of ${N##0}:

echo ${N#0} # removes only one leading zero
echo ${N##0} # removes all leading zeroes

but in my case, that doesn't work:

${N##0} works as ${N#0}, i.e, only one leading zero is removed in both cases.

Can anyone help me ?

P.D: I'd prefer this shell trick rather than using an external command.

Re: Parameter substitution in shell: does ${##} work ?

Peter Godron — Tue, 24 Oct 2006 06:09:44 GMT

Hi,
N="000034"
N2=`expr $N + 0`
echo $N2
34

Re: Parameter substitution in shell: does ${##} work ?

Peter Godron — Tue, 24 Oct 2006 06:13:22 GMT

Hi (again),
also found:
http://forums1.itrc.hp.com/service/forums/questionanswer.do?threadId=61952

Re: Parameter substitution in shell: does ${##} work ?

spex — Tue, 24 Oct 2006 06:43:39 GMT

Hi,

$ N=000034
$ echo ${N##*0}
34

For more information on parameter substitution:
http://docs.hp.com/en/B2355-90046/ch19s03.html#d0e17999

PCS

Re: Parameter substitution in shell: does ${##} work ?

James R. Ferguson — Tue, 24 Oct 2006 06:50:38 GMT

Hi:

# N=000034
# X=${N##*0}
# echo $X

Note the splat ("*") following the "##".

Regards!

...JRF...

Re: Parameter substitution in shell: does ${##} work ?

Frank de Vries — Tue, 24 Oct 2006 07:01:31 GMT

Hello

Indeed right your are:
use typeset -i N=000034

It seems without interger the '*' but '?'
comes to the rescue :
see examples:

[vwbsrv3:/d]# print ${N##*0}
34
[vwbsrv3:/]# print ${N##?0}
0034
[vwbsrv3:/]# print ${N##??0}
034
[vwbsrv3:/]# print ${N#??0}
034
-No difference here

[vwbsrv3:/]# print ${N##???0}
34
[vwbsrv3:/stand/build]#

But when you make it a INTEGER,

[vwbsrv3:/stand/build]# typeset -i N=00034
[vwbsrv3:/stand/build]# print ${N##*0}
34
[vwbsrv3:/stand/build]# print ${N##0}
34

It works:

Re: Parameter substitution in shell: does ${##} work ?

CAS_2 — Tue, 24 Oct 2006 07:31:58 GMT

James, read these examples below:

example 1: OK

X=00000034
echo ${X##*0}
34

example 2: failed

X=50034
echo ${X##*0}
34

Re: Parameter substitution in shell: does ${##} work ?

James R. Ferguson — Tue, 24 Oct 2006 07:50:53 GMT

Hi (again) Cas:

Ok, that's obvious in retrospect. We haven't nor can't anchor to the beginning of the string. I hadn't paid attention to that.

Using 'typeset' keeps things shell-bound:

# typeset -LZ N=0000034;echo ${N}
34

# typeset -LZ N=1000034;echo ${N}
1000034

Regards!

...JRF...

Re: Parameter substitution in shell: does ${##} work ?

spex — Tue, 24 Oct 2006 08:04:15 GMT

Hello,

# N1=0000034
# N2=50034
# echo $N1
0000034
# echo $N2
50034
# echo $(( 10#$N1 ))
34
# echo $(( 10#$N2 ))
50034

PCS

Re: Parameter substitution in shell: does ${##} work ?

Peter Nikitka — Tue, 24 Oct 2006 08:38:17 GMT

Hi,

if you really cannot use a typeset because of restrictions in the value, your variables are set, this will do it:

X=00000034
print ${X#${X%%[1-9]*}}
34

X=00010034
print ${X#${X%%[1-9]*}}
10034

X=500000034
print ${X#${X%%[1-9]*}}
500000034

The trick is, to determine the number of leading zeros (if any exist) first.

mfG Peter

Re: Parameter substitution in shell: does ${##} work ?

Arturo Galbiati — Wed, 25 Oct 2006 03:03:33 GMT

Hi,

/> X=00000034
/> echo $X
00000034
/> X=$(( X+0 ))
/> echo $X

the rtick is to add 0 to force the varibale to be an integer. In this way the leading zeros will be removed.

HTH,
Art

Re: Parameter substitution in shell: does ${##} work ?

CAS_2 — Wed, 25 Oct 2006 03:25:10 GMT

Thanks, Peter