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тАО10-23-2006 10:57 PM
тАО10-23-2006 10:57 PM
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.
Solved! Go to Solution.
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тАО10-23-2006 11:09 PM
тАО10-23-2006 11:09 PM
Re: Parameter substitution in shell: does ${##} work ?
N="000034"
N2=`expr $N + 0`
echo $N2
34
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тАО10-23-2006 11:13 PM
тАО10-23-2006 11:13 PM
Re: Parameter substitution in shell: does ${##} work ?
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тАО10-23-2006 11:43 PM
тАО10-23-2006 11:43 PM
Re: Parameter substitution in shell: does ${##} work ?
$ N=000034
$ echo ${N##*0}
34
For more information on parameter substitution:
http://docs.hp.com/en/B2355-90046/ch19s03.html#d0e17999
PCS
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тАО10-23-2006 11:50 PM
тАО10-23-2006 11:50 PM
Re: Parameter substitution in shell: does ${##} work ?
# N=000034
# X=${N##*0}
# echo $X
Note the splat ("*") following the "##".
Regards!
...JRF...
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тАО10-24-2006 12:01 AM
тАО10-24-2006 12:01 AM
Re: Parameter substitution in shell: does ${##} work ?
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:
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тАО10-24-2006 12:31 AM
тАО10-24-2006 12:31 AM
Re: Parameter substitution in shell: does ${##} work ?
example 1: OK
X=00000034
echo ${X##*0}
34
example 2: failed
X=50034
echo ${X##*0}
34
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тАО10-24-2006 12:50 AM
тАО10-24-2006 12:50 AM
Re: Parameter substitution in shell: does ${##} work ?
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...
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тАО10-24-2006 01:04 AM
тАО10-24-2006 01:04 AM
Re: Parameter substitution in shell: does ${##} work ?
# N1=0000034
# N2=50034
# echo $N1
0000034
# echo $N2
50034
# echo $(( 10#$N1 ))
34
# echo $(( 10#$N2 ))
50034
PCS
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тАО10-24-2006 01:38 AM
тАО10-24-2006 01:38 AM
Solutionif 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