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If I have the following $STRING

aaa.bbb.ccc.[ddd].eee.fff.[ggg].hhh

is there any way, using bash parameter expansion, to echo the following

aaa.bbb..ccc.eee.fff..hhh

That is, remove all occurrences of square brackets and everything inside those brackets? Everything I've tried ends up either removing everything in the string after the first left bracket or removing the brackets but leaving behind everything inside the brackets.

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1  
Could you verify the output in this post again? –  Ivan Chau Dec 10 '13 at 9:44

2 Answers 2

up vote 1 down vote

Requires shopt -s extglob:

bash-4.1# STRING='aaa.bbb.ccc.[ddd].eee.fff.[ggg].hhh'

bash-4.1# echo "${STRING//\[+([^\]])\]}"
aaa.bbb.ccc..eee.fff..hhh
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up vote 1 down vote 
shopt -s extglob

STRING='[asd].aaa.bbb.ccc.[ddd].eee.fff.[ggg].hhh.[asd]'
printf '%s\n' "${STRING//@(.\[*([^]])]|\[*([^]])]?(.))/}"

would give:

aaa.bbb.ccc.eee.fff.hhh
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