aaaaaaaa 09 bbbbbbbb 90 ccccccccccccccc 89 ddddd 09 Using sed/awk/replace, in the above text I want to remove anything that comes after the first space in each line. For example the output will be:
aaaaaaaa bbbbbbbb ccccccccccccccc ddddd any help will be appreciated.
Sed
sed 's/\s.*$//' Grep
grep -o '^\S*' Awk
awk '{print $1}' As pointed out in the comments, -o isn't POSIX; however both GNU and BSD have it, so it should work for most people.
Also, \s/\S may not be on all systems, if yours doesn't recognize it you can use a literal space, or if you want space and tab, those in a bracket expression ([...]), or the [[:blank:]] character class (note that strictly speaking \s is equivalent to [[:space:]] and includes vertical spacing characters as well like CR, LF or VT which you probably don't care about).
The awk one assumes the lines don't start with a blank character.
cut -d ' ' -f 1 < your-file would be the most efficient.
cut, I can't tell if it will accept any sort of regex syntax. Do you have any idea whether or not it will, and if so, which type(s)? sry to bother and thanks. awk '{print $1}' file or
sed 's/ .*//' And the one through perl,
$ perl -pe 's/^([^ ]+) .*$/\1/' file aaaaaaaa bbbbbbbb ccccccccccccccc ddddd Through GNU grep,
$ grep -oP '^[^ ]*' file bbbbbbbb ccccccccccccccc ddddd 