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
-Derek
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