UNIX for Dummies Questions & Answers

Hi Everyone,

Someone please tell me how to use grep with the -f option.

I need to search for a particular word and display the line that has that word. I need to do this on multiple files so I thought I'll create a file which lists the filenames that need to be searched and have grep go through the list in the file, but its not working.

Lets say I'm looking for the word "ABC" in the file POT1, POT2 & POT3

I would do "grep -w ABC POT1 POT2 POT3" and that would list the lines containing the word ABC.

I also need to list the files that does not contain the search word, do I use -v option?

I will be putting this command into a script so if anyone have a better way of achieving this please do share it with me.

Thank you so much.

Try something like:

Code:

grep 'ABC' POT*

Thanks Franklin52

The problem is that it searches all the files that begin with POT and there are hundreds. I'm only interested in files that were created for the day.

So start by excluding the other cases:
e.g

Code:

grep 'ABC' $(ls -l POT*|grep " Aug 17 "|cut -c 58-80)

Thanks vbe, that command worked. Now, how do I find out which of those files does not contain the search word. I just want to display the file names.

I have been trying the grep command with the -v option but without any success.

-v is for rejecting e.g.

Code:

ps -ef | grep vbe| grep -v grep # Im not interested by the grep process...

In maths you would have:
$(ls -l POT*|grep " Aug 17 "|cut -c 58-80) = A
grep 'ABC' $(ls -l POT*|grep " Aug 17 "|cut -c 58-80) = B

So

A - B = C
C is whats left: What was not retained by grep ...

Code:

find /YOUR_PATH -name "POT*" -type f -mtime -1 -exec grep "ABC" {} \;

---------- Post updated at 11:14 AM ---------- Previous update was at 11:12 AM ----------

use -l option to display filename only.

Code:

grep -v "ABC" -l POT*