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ahmedb72

07-13-2009 01:29 AM

About trimming a variable

Hi,
When I studied variable or parameter expansions, I learned the following:

${variable%pattern} trim the shortest match from the end
${variable%%pattern} trim the longest match from the end
I couldn't figure out the difference between them.

Code:

#var1=1100

#echo ${var1%0}

110

#echo ${var1%%0}

110

Thanks in advance.

Disillusionist

07-13-2009 01:59 AM

Try:

echo ${var1%0*}

echo ${var1%%0*}

ahmedb72

07-13-2009 05:08 AM

Hi Disillusionist,
Your code example illustrated for me the difference but it violated the trimming purpose. I simply want to trim the leading zeros. eg: 1100 -> 11 , 210 -> 21

Code:

#var1=11001

#echo ${var1%0*}

110                    <--- no trimming

#echo ${var1%%0*}

11

colucix

07-13-2009 05:36 AM

The wildcard * has the bash meaning, not the regexp one. In parameter substitution

Code:

${var%%0*}

it means any character from zero to the end of the string. In your first example, instead, the pattern matched just one zero, so that using either %% or % the result is the same. If you want to remove any number of trailing zeros you have to activate the extglob option:

Code:

$ var=1100

$ shopt -s extglob; echo ${var%%+(0)}; shopt -u extglob

See man bash under the section pathname expansion --> pattern matching for details.

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