[linux] How can I use grep to show just filenames on Linux?

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.

This question is related to linux grep

The answer is


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.


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 (+).


Your question How can I just get the file-names (with paths)

Your syntax example find . -iname "*php" -exec grep -H myString {} \;

My Command suggestion

sudo find /home -name *.php

The output from this command on my Linux OS:

compose-sample-3/html/mail/contact_me.php

As you require the filename with path, enjoy!


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.)