linux,string,bash,unix,awk , How to remove characters before and including an underscore?


How to remove characters before and including an underscore?

Question:

Tag: linux,string,bash,unix,awk

In bash scripting what's an efficient way to do the following please?

var="fooo_barrrr"

What is the best way to remove all characters before and including the '_' so that var becomes "barrrr" please?


Answer:

Using Parameter Expansion:

$ var="fooo_barrrr"
$ echo ${var#*_}
barrrr

To change the var itself, var=${var#*_}.

Note this removes up to the first _:

$ var="fooo_barrr_r"
$ echo ${var#*_}
barrr_r

If you wanted to remove up to the last one, you would need to use ## instead:

$ var="fooo_barrr_r"
$ echo ${var##*_}
r

See some alternatives:

With sed:

$ sed 's/^[^_]*_//' <<< "foo_barrrr_r"
barrrr_r

With awk:

$ awk 'gsub(/^[^_]*_/,"")1' <<< "foo_barrrr_r"
barrrr_r

With grep:

$ grep -oP '[^_]*_\K.*' <<< "foo_barrrr_r"
barrrr_r

In all cases, to store the new value in the same var, do var=$(command <<< "$var").


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