UNIX for Dummies Questions & Answers

This suggested command contains design and implementation errors. Consider what happens on April 1st. The grep match string becomes "Apr01" which will not match a directory list.


A less-than-perfect way of picking up files and directories created today is to match against the date and check that there is a colon character somewhere on the line.

Code:

TODAY="`date '+%b %e'`"        # Mmm d
ls -lar | grep "${TODAY}"|grep ":"

The importance here is to get the "date" command right and to preserve spaces in the date throughout by proper use of double quote characters.

Beware of dates with a single digit day. The command "date +%d" which is wrongly advised in various earlier posts gives the day of the month with a leading zero but "ls -lar" give the day of the month with a leading space. Therefore we must use "date +%e" and preserve that space character.
My example is not a recommended method (mainly because it does not work with filenames containing colon characters and is awkward with filenames containing space characters). My earlier post in this thread is a recommended method.

Dear experts,

Do you find find with touch to be much more reliable in scripts ?

If you are referring to post #1 , yes. This is mainly because the format of "ls -la" changes for older files (the time gets replaced with the year).
There are more techniques available depending on exactly what Operating System and tools you have.
For the moment we'll assume that you don't let an open log run through midnight. This can make scripting "on a daily basis" difficult.

Methyl the code you provided picks the files created on a particular day ie today and lists them on the screen, but my question is, after you run this script, lets say i have 2 files created on the same day, How do i copy these 2 files to a different location..if this is answered then i believe i can close this discussion for good..Please help..Thanks

Well, you cannot hide ' inside ', and you have to do a 'command stdout to string' conversion $(...) or `...`. Think of echo as 'command line string to stdout' converter, and this is the opposite.

Code:

cp $(
 ls -lrt | awk '
   {print $6$7" "$9 }
  ' | grep `date +%b""%d` | awk '
   {print $2}
  ' ) somewhere/new

Now, if you dislike nesting:

Code:

 ls -lrt | awk '
   {print $6$7" "$9 }
  ' | grep `date +%b""%d` | awk '
   {print $2}
  ' | mv $( cat ) somewhere/new

Both of these are ugly if no files, so sometimes, put the list in an env. var.:

Code:

new=$(
 ls -lrt | awk '
   {print $6$7" "$9 }
  ' | grep `date +%b""%d` | awk '
   {print $2}
  '
 )
 
if [ "$new" != "" ]
then
 mv $new somewhere/new
fi

If the number of files is really big, like from a long find, then waiting for a complete list is an unconscionable delay, so you want to pipeline parallel more:

Code:

ls -lrt | awk '
   {print $6$7" "$9 }
  ' | grep `date +%b""%d` | awk '
   {print $2}
  ' | while read f
 do
  mv $f somewhere/new
 done

but this lacks economy of scale on mv calls (nice for must-do-one-at-a-time commands), so use xargs to batch them up:

Code:

ls -lrt | awk '
   {print $6$7" "$9 }
  ' | grep `date +%b""%d` | awk '
   {print $2}
  ' | xargs -n101 echo | while read fs
 do
  mv $fs somewhere/new
 done

xargs is 'pile args from this many stdin lines on the end of the supplied command and execute'. The number 101 is magic, in that you can argue that you got over 99% of any economy of scale while minimizing latency to first mv. Specifying the number to xargs with -n not only limits the delay to first mv, it also says no execute for 0 lines.

The grep pattern needs to be `date '+^%b''%d '` so it both matches the date only at beginning of line (^) not in the file name, and matches the space separator (May1 and not May19 ).

This is just about perfect. Awesome piece of work Pickett..Thanks a ton ..You are the Man

Good tricks to use for lots of stuff! Enjoy!