Gary Kline wrote:
On Wed, Apr 25, 2007 at 06:21:52AM -0500, Derek Ragona wrote:
At 02:29 AM 4/25/2007, Gary Kline wrote:
       Guys,

       This is an awk-type question.  Hopefully a one-liner.  If I
       need to use #!/usr/bin/awk and a BEGIN/END (or whatever it is),
       that's okay...

       I want to do an ls -l in a  /home/kline/<directory> and find and
       edit files that are dated (let's say) Apr 19 or Mar 26.  This
       works to print $9 the filenames.

       ls -l| awk '{if ($6 == "Apr" && $7 == 19  || $6 == "Mar" && $7
       == 26 ) print $9}'

       What's the final part to get awk to vi $9?  Or another pipe and
       xargs and <what> "vi"?  Nothing simple works, so thanks for any
       clues!
I would use a simple approach incase you need to re-edit the list since editing will change file times: ls -l| awk '{if ($6 == "Apr" && $7 == 19 || $6 == "Mar" && $7 == 26 ) print $9}' > /tmp/myfilelist
then you can:
for i in `cat /tmp/myfilelist`;do vi $i;done

if you don't want to use a file, you can do in one shell loop too, but again this will change your file modification times: for i in `ls -l| awk '{if ($6 == "Apr" && $7 == 19 || $6 == "Mar" && $7 == 26 ) print $9}'`;do vi $i;done


        Yep; this is the simple kind of script I had in mind first but
wasn't sure if/how it would work. Your one-liner works "as-advertized", but then as you note, the timestamp is
        changed!! (duh)...  So it does make more sense to put the list
        into a /tmp/<foo> file.  Save typing when I re-edit.

        thanks much, indeed,

        gary


        -Derek

Don't forget my friendly, friend cut(1) (almost forgot that in my previous post). I think it's a lot more lightweight and faster than awk is; the only drawback is that delimiters are only 1 character wide, whereas heavier weight text processing tools can do multiple character search and replacements (sed, awk, perl, etc).

ls -l | cut -d ' ' -f 9 | xargs vi {} \; # change -f to meet your needs

-Garrett
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