Remove leading digits from a string with Bash using parameter expansion

Question

The initial string is RU="903B/100ms" from which I wish to obtain B/100ms.

Currently, I have written:

#!/bin/bash
RU="903B/100ms"
RU=${RU#*[^0-9]}
echo $RU

which returns /100ms since the parameter expansion removes up to and including the first non-numeric character. I would like to keep the first non-numeric character in this case. How would I do this by amending the above text?

score 2 · Answer 1 · answered Aug 11 '16 at 11:04

2

Assuming shopt -s extglob:

RU="${RU##+([0-9])}"

answered Aug 11 '16 at 11:04

melpomene

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fedorqui · Accepted Answer · 2016-08-11T11:11:01.507

2

You can use BASH_REMATCH to extract the desired matching value:

$ RU="903B/100ms"
$ [[ $RU =~ ^([[:digit:]]+)(.*) ]] && echo ${BASH_REMATCH[2]}
B/100ms

Or just catch the desired part as:

$ [[ $RU =~ ^[[:digit:]]+(.*) ]] && echo ${BASH_REMATCH[1]}
B/100ms

edited Aug 11 '16 at 11:11

answered Aug 11 '16 at 11:09

fedorqui

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Thank you! Would up vote but according to stackoverflow my gratitude is temporarily meaningless since I'm new here ;) – AASJC Aug 11 '16 at 11:53
@AASJC your kind words are way more meaningful than some virtual points :) – fedorqui Aug 11 '16 at 11:55

score 0 · Answer 3 · answered Aug 11 '16 at 10:59

0

echo "903B/100ms" | sed 's/^[0-9]*//g'
B/100ms

answered Aug 11 '16 at 10:59

P....

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Thanks, but preferably by using parameter expansion? This is a special case of a more general task I'm doing – AASJC Aug 11 '16 at 11:01

Remove leading digits from a string with Bash using parameter expansion

3 Answers3