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How can I use grep to show just file-names (no in-line matches) on Linux?
I am usually using something like:
find . -iname "*php" -exec grep -H myString {} \;
How can I just get the file-names (with paths), but without the matches? Do I have to use xargs? I didn't see a way to do this on my grep man page.
vim $(grep -rl Something app/)
grep -lr 'text-to-find' ./* works quite nicely!
The standard option grep -l (that is a lowercase L) could do this.
From the Unix standard:
-l
(The letter ell.) Write only the names of files containing selected
lines to standard output. Pathnames are written once per file searched.
If the standard input is searched, a pathname of (standard input) will
be written, in the POSIX locale. In other locales, standard input may be
replaced by something more appropriate in those locales.
You also do not need -H in this case.
From the grep(1) man page:
-l, --files-with-matches Suppress normal output; instead print the name of each input file from which output would normally have been printed. The scanning will stop on the first match. (-l is specified by POSIX.)
For a simple file search, you could use grep's -l and -r options:
grep -rl "mystring"
All the search is done by grep. Of course, if you need to select files on some other parameter, find is the correct solution:
find . -iname "*.php" -execdir grep -l "mystring" {} +
The execdir option builds each grep command per each directory, and concatenates filenames into only one command (+).
My command suggestion for getting the filename with path
sudo find /home -name *.php
The output from this command on my Linux OS:
compose-sample-3/html/mail/contact_me.php
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grep -nrl "some text" .when looking for text in a set of subfolders recursively