UNIX for Dummies Questions & Answers

I tried the code above but it did not list filenames as intended instead it output the contents of all the files that did not have the search word.

There should be better way to search them, but here is a solution.

Code:

diff <(ls POT*) <(grep -l "ABC" POT*) |awk '/^</ {print $2}'

Why not use ctime? Also, do you know the difference with using -daystart and not? I want to know 'cause I think it's really required if what the user means today and not 24 hours before.

---------- Post updated at 03:43 PM ---------- Previous update was at 03:36 PM ----------

So do you mean you want to output the files that do have he search word? Does not not using -v will make the solution?

Code:

grep -l ABC POT*

Edit: Or maybe not. But the former code seems to work fine.

Code:

grep -vl ABC POT*

You can also try to add the option -F or use fgrep

Code:

grep -F -vl ABC POT*
fgrep -vl ABC POT*

Hi rdcwayx,

I tried your command and it gave me a syntax error message. I typed to the commands exactly as you have given them, and I checked it twice for any typo error.



---------- Post updated at 00:57 ---------- Previous update was at 00:49 ----------

Hi konsolebox,

I tried all your commands above and all of them are listing the filenames only regardless of weather they contain the search word or not.

The time does not matter because the files are created at exactly at 3 pm everyday and I only want to do the search on those files that were created for the day.

Basically I just want to know which of those files contain the word I'm looking for and which does not.

I run in bash, if you met problem, you can run it separately

Code:

ls POT* > temp1
grep -l "ABC" POT* > temp2
diff temp1 temp2 |awk '/^</ {print $2}'

Hi rdcwayx,

Code:

ls POT* > temp1
grep -l "ABC" POT* > temp2
diff temp1 temp2 |awk '/^</ {print $2}'

the above code worked. If only I could have it search the files that were created for the day, that would be perfect.