I need your help.

LEO111 · #1 (**permalink**) 09-15-06, 21:11

Hi
I have a multiple files in the same directory . The only different between these file is the date stamp.
For example : a_20060308.dat
a_20060309.dat

without the datestamp, these files will have the same file name (which is a)
and of course, these two have different created date.

I just want to pickup the lastest file and mv that latest file to different location. Does anyone knows or have any idea how to get this using korn shell or awk/sed ...? Thank you so much in advance.

Tyveleyn · #2 (**permalink**) 09-16-06, 06:18

If you want to do it by timestamp you can write:

Quote:

mv $(ls -ltr a_*.dat | awk 'END{print $9}') somelocation

where the '$9' in the awk statement represents the ninth (last) field in the ls -l outputline. This is the filename on many UNIX / Linux systems. By using ls -ltr the listing is sorted on t(ime)r(eversely).

Regards

BTW. This is Linux syntax, if it doesn't work on your system you always can use `` instead of $().

LEO111 · #3 (**permalink**) 09-16-06, 18:45

TYVELEYN

Thank you so much for your help regarding this matter. I'll try this approach and will let you updated. Again, Thank you and Have a nice weekends.

Best Regards,
LEO111.

LEO111 · #4 (**permalink**) 09-16-06, 18:49

TYVELLEYN
The files name is only example... The problem is I have lot of file name and I don't want to specify in my scripts what is the name of the files. Whatever files are there , I just want to pick up the lastest one if duplicate. So I can not hard coded the file name there. Is there a way to get rid of the datestamp first then apply the ls -ltr command to it to get the lastest file. Thank you.

pdreyer · #5 (**permalink**) 09-18-06, 03:08

e.g.
latestfile=$(ls -A1t /tmp |head -1)

What do you mean by "if duplicate"
There can't be duplicate filenames in a directory
Do you want to sort on file name?
Just use ls -A1r instead

Tyveleyn · #6 (**permalink**) 09-18-06, 04:33

Maybe this is what you're looking for:

Code:

#!/usr/bin/ksh                                                                
                                                                              
for i in $(ls -tr *[12][09][0-9][0-9][01][0-9][0-3][0-9]*); do
    cp $i somelocation/$(echo $i | sed 's/^\(.*[^0-9]\)[12][09][0-9][0-9][01][0-9][0-3][0-9]\([^0-9].*\)$/\1\2/')
done

This is a very basic algorithm for detecting files with a datevalue in their names (yyyymmdd). It cuts out the datevalue from every filename found and writes the original files to a file with the stripped name.
E.g. every a_YYYYMMDD.dat file will be written to 'a_.dat' and every b_YYYYMMDD.dat file to 'b_.dat'. Because the filenames are fetched with 'ls -tr' (adopted from pdreyer, didn't know it would work without the long list...) the files are reversely sorted by timestamp and thus overwrite the previous (= older) file with the target filename.

Regards

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