grep a text in files and print the file name who don't contain such text

whossa · 04-12-2012, 08:37 AM

Hello all,

I'm going to grep a text in the files and wanna to just print the file names who don't contain this text, apparently, use
find . -name ***|xargs grep -lv pattern won't work, this will print all files.

Then I use
find . -name ***|tee ../all.txt|xargs grep -l pattern|tee ../contained.txt

then
diff ../all.txt ../contained.txt

It could work yet too stupid, do anyone have elegant solution? Much appreciated! I could use perl to filter out this two file list yet this may be unnecessary!

MensaWater · 04-12-2012, 09:03 AM

grep -L

From the man page:

Quote:

-L, --files-without-match
Suppress normal output; instead print the name of each input
file from which no output would normally have been printed. The
scanning will stop on the first match.

For most UNIX/Linux commands there are manual (man pages). Typing "man <command>" will show it to you. Also on Linux some things also have info pages. Typing "info <command>" will show that to you. You can even type "man man" to find out how the man command works.

pan64 · 04-12-2012, 09:05 AM

grep -lc will print counts, and you can grep the zeros:
grep -lc pattern files | grep ':0$'

from the other hand in general instead of the chain of find, tee, xargs, grep and tee (and maybe sed, awk, and cat) you would better try a single perl script.

MensaWater · 04-12-2012, 09:22 AM

UNIX/Linux is case sensitive. "-l" and "-L" have different meanings - in fact they do the exact opposite of each other.

whossa · 04-12-2012, 09:07 PM

Thanks for MensaWater's help, that's great!

MensaWater · 04-13-2012, 07:49 AM

Glad I could help.

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