Grep From Files and Display the File Name

Posted on in Categories last updated February 18, 2011

How do I grep from a number of files and display the file name only?

When there is more than one file to search it will display file name by default:

grep "word" filename
grep root /etc/*

Sample outputs:

/etc/bash.bashrc:       See "man sudo_root" for details.
/etc/crontab:17 *       * * *   root    cd / && run-parts --report /etc/cron.hourly
/etc/crontab:25 6       * * *   root    test -x /usr/sbin/anacron || ( cd / && run-parts --report /etc/cron.daily )
/etc/crontab:47 6       * * 7   root    test -x /usr/sbin/anacron || ( cd / && run-parts --report /etc/cron.weekly )
/etc/crontab:52 6       1 * *   root    test -x /usr/sbin/anacron || ( cd / && run-parts --report /etc/cron.monthly )
grep: /etc/gshadow: Permission denied
/etc/logrotate.conf:    create 0664 root utmp
/etc/logrotate.conf:    create 0660 root utmp

The first name is file name (e.g., /etc/crontab, /etc/group). The -l option will only print filename if th

grep -l "string" filename
grep -l root /etc/*

Sample outputs:

grep: /etc/at.deny: Permission denied

You can suppress normal output; instead print the name of each input file from which no output would normally have been printed:

grep -L "word" filename
grep -L root /etc/*

Sample outputs:


13 comment

    1. Why ls? It will fail when you have file names with white spaces. A better way is to use wild cards:

      for i in * ; do
      grep -q “whatever” $i && echo $i
      1. what about
        grep -rI “whatever” ./ | cut -d: -f1 | sort -u

        nb it will fail on file containing “:” such as some perl man pages but…

        …Or just add some proper form
        for i in `ls` ; do
        grep -q “whatever” “$i” && echo “$i”

  1. or if you need to search all file and subdirs within a dir you can run the commands above. Just replace `ls` with `find . -type f`.

  2. Or better yet, use xargs.
    find . -type f | xargs grep html

    If you want to deal with spaces etc…
    find . -type f | xargs -I {} grep -H foo "{}"

  3. How do I grep if i am looking for a specfic percentage?
    grep -w “68%-100% /usr/local/stage”

    thanks in advance

  4. how do I know the location and file name form the result that grep shows?
    I use command like cat * | grep
    Now what I need is the location of the file name along withe the grep result.

    Help anyone

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