mjhjmtu

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 'BEGINFS="\> +" print $1'
a b

but unfortunately not the full-blown ones from perl or pcre (eg. (?<!\), (?<=w), etc):

$ echo 'a b, c' | perl -nle '@a=split /(?<!\)s+/, $_; print $a[0]'
a b,

You could substitute space with something else and back again afterwards.

sed 's/\ /\x20/g' data_file | awk ' print $1; ' | sed 's/\x20/\ /g'

With just sed:

sed -r 's/^((([^]*\ )1,)?[^ ]*).*/1/' infile

Or shorter:

sed -r 's/^(([^]*\ )*[^ ]*).*/1/' infile

This (([^]*\ )1,)?[^ ]* matches:

• 
  [^]*\: anything that it's not a back-slash which ends with back-slash followed by a space (note that inside character class is not required to be escaped, but outside does).


• 
  ([^]*\ )1,: matching above with one-or-more times of occurrences.


• 
  (([^]*\ )1,)?: this is optional when using (...)?; we could use ([^]*\ )0, instead as well or ([^]*\ )*.


• 
  ((([^]*\ )1,)?[^ ]*): matches above which is optional followed by anything that it's not a space and hold as group match with 1 as its back-reference.


• 
  ((([^]*\ )1,)?[^ ]*).*: matches above (...) and anything else .*.

then is replacement part just print the 1 which is the output:

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 grep or compatible:

grep -Po '^(\.|S)*'

Or with ERE:

grep -Eo '^(\.|[^[:space:]])*'

That treats as a quoting operator, for whitespace as a delimiter, but also for itself. That is, on foo\ bar input, it returns foo\.