How can I add space after filename of grep output?

Question

I am using grep to search matching patterns inside files, and copy its filename from the output. Since I do this operation a lot I may end up copying : as well. Hence is it possible to add two space before the first matching semicolumn or ending of the filename?

$ grep -Hrn -e "zorro" --color=always --exclude-dir=dir -I -F . | \
    cut -c 11- | awk '$0="\033[33;35m"$0' | tr -s '[:space:]'
doo.tex:1:zorro alper alper alper skfjsdlkfj dslfj dslkj

wanted output:

doo.tex  :1:zorro alper alper alper skfjsdlkfj dslfj dslkj
        ^
  <two space added>

Answer 1

My first though went to awk

awk -F: -vOFS=: '{$1 = $1 "  "; print}'

But that's doing more work than needed I think, splitting and re-joining the line.

The sed alternative is quite concise

sed 's/:/  :/'

replacing the first colon with 2 spaces and a colon.

---

Regarding the filenames with colons conversation:

Reading the GNU grep man page: use the -Z option: "Output a zero byte (the ASCII NUL character) instead of the character that normally follows a file name."

So:

# .......v
grep -HrnZ -e "zorro" --color=always --exclude-dir=dir -I -F . | \
  perl -pe 's/\0/  :/'

Answer 2

If all you need is a list of filenames containing text that matches a pattern, such as "zorro", and you intend to use those filenames with another program, then you can do something like the following to create a bash array containing the filenames, and avoid a lot of manual copy-pasting with the mouse.:

mapfile -d '' -t myfiles < <(grep -IFlZr zorro ./)

mapfile is a bash built-in command that populates an array from stdin. In this case, stdin is coming from grep running as a process substitution. I'm using -d '' with mapfile to tell it the input will be delimited by NUL characters, and -Z with grep so that it outputs a NUL-separated list of matching filenames. This will work with any filename, no matter what characters are in the name - including colons, spaces, newlines, and shell metacharacters.

You can see what's in the array with declare -p myfiles, or use the array's elements as args to another program (e.g. printf '%s\'n' "${myfiles[@]}"), or in a loop like for f in "${myfiles[@]}"; do echo "$f" ; done.

Note: grep's -l option will exit after the first match (same as the -m option), which will speed up the search in files where the search string appears early in the file. In files where the pattern doesn't appear, it still has to read the entire file.

BTW, if the reason why you're using the -I option is that you want to avoid binary files (like .pdf or .dvi output from TeX), then you can use find instead of grep -r. For example:

mapfile -d '' -t myfiles < <(find . -type f -name '*.tex' -exec grep -lZF zorro {} +)

---

If you need more complicated selection criteria, beyond what grep is capable of then you can use awk or perl or whatever instead of grep. e.g. if you wanted only filenames where "zorro" appeared on line 3 of the file:

mapfile -d '' -t myfiles < <(find . -type f -name '*.tex' \
  -exec perl -n -e 'if ($. == 3) {
                      printf "%s\0", $ARGV if (/\Qzorro\E/);
                      close(ARGV);
                    }' {} +)

This is just a simple example - if you can think of a criterion and write it as a perl (or awk or python or whatever) script, then you can use it to selectively populate the array. You can use any command, or any long and complicated pipeline of commands, as long as it prints filename(s) to stdout separated by a NUL character.

Note: the close(ARGV) close the current file and skips to the next, if any, whether the filename was printed or not. This means it never has to read more than three lines of any file, so will be a lot faster if you have a lot of large files to search.

Note 2: the \Q and \E is perl's way of doing a fixed-string search, similar to -F with grep. If you were using awk, you could do something similar with the index() function.