Remove directory path from grep -r output

Question

I am running grep -r to look for the context of a word in multiple files.

I am using -r to do it recursively, -i to ignore case and -C to get lines below and above:

grep -r -i -C 10 --group-separator="==========" "29/04/2020" "$dir" >> output.txt

In my output, however, I get the filenames before the match, like:

../data/filename1.txt-       (other text)
../data/filename1.txt-        29/04/2020 is the date for etc
../data/filename1.txt-       (other text)                                                                                      
==========
../data/different_filename.txt-       (other text)
../data/different_filename.txt-        something in 29/04/2020
../data/different_filename.txt-       (other text) 

I would like as output just:

(other text)
29/04/2020 is the date for etc
(other text)                                                                                      
==========
(other text)
something in 29/04/2020
(other text) 

Do you know how I could alter the grep -r command to exclude the filepaths?

Answer 1

Use grep -h, as described in man grep:

-h
--no-filename
Suppress the prefixing of file names on output. This is the default when there is only one file (or only standard input) to search.

SEE ALSO:
Get grep to not output file name