How to remove initial and final digits from strings using bash scripting

Question

Quite new to bash scripting so If I have a list usually text files each file specific for one country for example USA.txt client addresses with leading ZIP and ending P.O Box (Numbers are masked for privacy) :

38495newyork940 454354irvine40404 2333wathington4440 32333california002929 23234dallas4th93303

I want to run USA.txt as an argument to a bash script that would remove the leading and final digits from each line. Please note that certain cities or states might have numbers within the names so I want to keep them.

I tried to run it against sed commands but I can't combine the end result I just remove the leading or final numbers with commands like :

for leading digits:

sed -r 's/^[0-9]+(.*[^0-9].*)$/\1/g'

and for ending digits:

sed "s/\([^0-9]\)[0-9]*\s*$/\1/"

I need the output like:

newyork irvine wathington california dallas4th

Thanks

Lev Levitsky · Accepted Answer · 2017-07-29 14:29:37Z

3

You can give sed as many commands as you like:

$ sed 's/^[0-9]*//; s/[0-9]*$//' USA.txt newyork irvine wathington california dallas4th

Or you can do it all in one command:

$ sed 's/^[0-9]*\(.*[^0-9]\)[0-9]*$/\1/' USA.txt newyork irvine wathington california dallas4th

answered Jul 29, 2017 at 14:29

Lev Levitsky

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2 Comments

Sundeep Over a year ago

and a 3rd way would be sed -E 's/^[0-9]*|[0-9]*$//g or sed -E 's/^[0-9]*|[0-9 ]*$//g' (OP's sample has spaces at end)

Lev Levitsky Over a year ago

@Sundeep The spaces seem to be an artifact. They are not there in the edit mode (except for the last sample line).

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