Issue
I find grep's --color=always flag to be tremendously useful. However, grep only prints lines with matches (unless you ask for context lines). Given that each line it prints has a match, the highlighting doesn't add as much capability as it could.
I'd really like to cat a file and see the entire file with the pattern matches highlighted.
Is there some way I can tell grep to print every line being read regardless of whether there's a match? I know I could write a script to run grep on every line of a file, but I was curious whether this was possible with standard grep.
Solution
Here are some ways to do it:
grep --color 'pattern\|$' file
grep --color -E 'pattern|$' file
egrep --color 'pattern|$' file
The | symbol is the OR operator. Either escape it using \ or tell grep that the search text has to be interpreted as regular expressions by adding -E or using the egrep command instead of grep.
The search text "pattern|$" is actually a trick, it will match lines that have pattern OR lines that have an end. Because all lines have an end, all lines are matched, but the end of a line isn't actually any characters, so it won't be colored.
To also pass the colored parts through pipes, e.g. towards less, provide the always parameter to --color:
grep --color=always 'pattern\|$' file | less -r
grep --color=always -E 'pattern|$' file | less -r
egrep --color=always 'pattern|$' file | less -r
Answered By - Ryan Oberoi Answer Checked By - Senaida (WPSolving Volunteer)