Shell Programming and Scripting

I have executed the below command:

find . -name "Ks*" -type f -exec ls -ltr {} \; | awk '{printf("%ld %s %d %s \n",$5,$6,$7,$8,$9)}'

and here is the output:

1282 Oct 7 2004
51590 Jul 10 2006
921 Oct 7 2004
1389 Jun 4 2003
1037 May 19 2004
334 Mar 24 2004
672 Jul 8 2003
977 Apr 1 2004

I want to display the name of the file and the directory where the file exists including the above output

Desired output:

same output + directory + filename

Thanks in advance...

Regards,
Venkat.

Specify the full path and you'll get the desired ouput:

Code:

find /full/path/here -name "Ks*" -type f -exec ls -ltr {} \; | awk '{printf("%ld %s %d %s \n",$5,$6,$7,$8,$9)}'

I think u missed a %s in your awk.

Thank you, but am getting a different output which i don't expect:

921 Oct 7 /opt/Scripts/KsGenerateSkuInvPix.sh
1389 Jun 4 /opt/Scripts/KsPurgeC_command_database.cfg

I just need the above output + directory in a separate column (no need to display the absolute path) + filename in a separate column

Hope i explained clearly.

Thanks,
Venkat.

for the following:

Code:

921 Oct 7 /opt/Scripts/KsGenerateSkuInvPix.sh

should the o/p be:

Code:

 
921 Oct 7 Scripts KsGenerateSkuInvPix.sh

Please tell if I ve followed you correctly!

not sure try this

Code:

find . -name "Ks*" -type f -exec ls -ltr {} \;|awk -F"[ /]" '{print $17" "$18" "$19" "$22" "$23}'

Thanks a lot for your help... but when i executed the above command , i just got blank output.

yes hkansal...

the desired output: 921 Oct 7 2004 Scripts KsGenerateSkuInvPix.sh