Delimit by space but ignore backslash space 5678 [] testing,\ group [] [testing [] ip\ 5.6.7.8 [] launch-wizard-1 0.0.0.0/0 456dlkjfa [] 1.2.3.4 [] test 1.2.3.4/32 4.3.2.0/23 4.3.2.0/23 default 4.3.2.0/23 4.3.2.0/23 launch-wizard-2 0.0.0.0/0 launch-wizard-3 0.0.0.0/0 2.3.4.5/32 [] I would like to get the first column of the above but the catch is that, I need to treat `\` **(backslash space)** as a part of the column, so `awk '{print $1}'` should give me 5678 testing,\ group [testing ip\ 5.6.7.8 launch-wizard-1 456dlkjfa 1.2.3.4 test default launch-wizard-2 launch-wizard-3 2.3.4.5/32

with gnu awk (`gawk`) you can use some zero-length assertions like `\<` or `\>`:


$ echo 'a\ b c' | gawk 'BEGIN{FS="\\> +"} {print $1}'
a\ b


but unfortunately not the full-blown ones from `perl` or `pcre` (eg. `(?

$ echo 'a\ b, c' | perl -nle '@a=split /(? a\ b,