I can do the following to see if some word is available in the output of "cat":
cat filename | grep word This filters the output and shows only those lines which contain "word". Now, is it possible to only highlight the "word" in the output, without dropping other lines?
You can grep for an EOL along with your real query (if you already have an alias for grep to use --color, as is default in many distributions, you can omit it in the following examples):
grep --color=auto 'word\|$' file Since the EOL is not a real character, it won't highlight anything, but it will match all lines.
If you would prefer not to have to escape the pipe character, you can use extended regular expressions:
grep -E --color=auto 'word|$' file 
GREP_COLOR(S) environment variable. gnu.org/software/grep/manual/html_node/… hl ("highlight") declared in my .zshrc. Something like cat foo.txt | hl bar. But it should also work when provided a file (hl "bar" foo.txt). If you haven't GNU grep available, here is something more portable:
grepc() { pattern=$1 shift esc=$(printf "\033") sed 's"'"$pattern"'"'$esc'[32m&'$esc'[0m"g' "$@" } You can customize the color using one of these codes
30m black 31m red 32m green 33m yellow 34m blue 35m magenta 36m cyan 37m white Using 7m instead of a color code will put the string in reverse video.
^[[32mword^[[0m ^[ is how the escape character is represented at least under vi, answer updated to avoid the ambiguity. cat testt.c | sed $'s/main/\E[31m&\E[0m/g' Hope it helps
grep word filenameis clearer and faster. ;) Also,lesscan do this while also providing paging. Just search forwordby typing/word(the term is actually a regular expression, just likegrep)