Shell Programming and Scripting

Hello,
I have a file packed with records similar to:

-rw-rw---- 1 pcpdsrv8 suppw 98737 21 Jan 09:08 Workflow_Calc_Distance.log.20101223060615.bin

Could someone tell me how I can cut out all the info up until the Date field please so it looks like:

21 Jan 09:08 Workflow_Calc_Distance.log.20101223060615.bin

I've tried various cut options but can't get it to work!

Cheers

Code:

sed -n 's/^.* \([0-3][0-9] .*$\)/\1/p' filename > result.out

Code:

awk '{print $6,$7,$8,$9}' filename > result.out

Code:

cut -d" " -f6-9 filename > result.out

Thanks Aia that's great. Only problem I have now is that it's taking the second line as a seperate record whereas it is in fact part of the first. The cut command now gives me:

21 Jan 09:08 Workflow_Calc_Distance.log.201

As the rest of the record is not on the same line. Is there an easy way to set a record or line length?

Cheers

Watch out for filenames with spaces (especially 2 or more together) as awk will strip them. Also look out for files older than 1 year as they have on time just date and year.

You are usually better of fetching the modification time for the files yourself (see gnu date -r option) rather than trying to cut them out of the ls output.