robby@nuchat.sccsi.com (Robert Oliver Jr.) (05/12/91)
Hi there,
Has anybody out there ever printed out the size,date and name of a file
without the permitions and owners just the size,date & name. I would like
to know how you did it!
Robert Oliver
robby@nuchat.sccsi.comtoma@swsrv1.cirr.com (Tom Armistead) (05/13/91)
In article <1991May12.085633.10066@nuchat.sccsi.com> robby@nuchat.sccsi.com (Robert Oliver Jr.) writes: >Hi there, > > Has anybody out there ever printed out the size,date and name of a file >without the permitions and owners just the size,date & name. I would like >to know how you did it! > > Robert Oliver > robby@nuchat.sccsi.com How about this? $ ls -l | awk '{ printf "%6s %s %2s %s %s\n",$5,$6,$7,$8,$9 }' Tom -- Tom Armistead - Software Services - 2918 Dukeswood Dr. - Garland, Tx 75040 =========================================================================== toma@swsrv1.cirr.com {egsner,letni,ozdaltx,void}!swsrv1!toma
phil@ux1.cso.uiuc.edu (Phil Howard KA9WGN) (05/13/91)
robby@nuchat.sccsi.com (Robert Oliver Jr.) writes: > Has anybody out there ever printed out the size,date and name of a file >without the permitions and owners just the size,date & name. I would like >to know how you did it! Assuming you want time with the date, you can pipe "ls -l" through this or some approximation of it to match your ls's output format: ----CUT-HERE--------CUT-HERE--------CUT-HERE--------CUT-HERE--------CUT-HERE---- #!/usr/local/bin/awk -f { if( NF>7 ) { z=substr(" ",length($4)) $4; z=z substr(" ",length($5)) $5; z=z substr(" ",length($6)) $6; z=z substr(" ",length($7)) $7; z=z " " $8; print z; }; }; ----CUT-HERE--------CUT-HERE--------CUT-HERE--------CUT-HERE--------CUT-HERE---- If you don't want the time, leave out the appropriate field. If you new awk is somewhere else, change the first line. Gawk may work, too, but I didn't try it. Watch out for some versions of ls that, when faced with a very high link count, will jam the link count number and the privileges together as one single field. To avoid that you might have to extract the files based on a combination of column number and whitespace, rather than whitespace alone. -- /***************************************************************************\ / Phil Howard -- KA9WGN -- phil@ux1.cso.uiuc.edu | Guns don't aim guns at \ \ Lietuva laisva -- Brivu Latviju -- Eesti vabaks | people; CRIMINALS do!! / \***************************************************************************/
dfpedro@uswnvg.UUCP (Donn Pedro) (05/16/91)
In article <1991May12.085633.10066@nuchat.sccsi.com>, robby@nuchat.sccsi.com (Robert Oliver Jr.) writes: > Hi there, > > Has anybody out there ever printed out the size,date and name of a file > without the permitions and owners just the size,date & name. I would like > to know how you did it! > > Robert Oliver > robby@nuchat.sccsi.com Did something like that just the other day. Here is the output of the ls -l command. $> ls -l -rw-r----- 1 dfpedro news 12956 May 8 22:47 Articles -rwx------ 1 dfpedro news 3626 Mar 13 14:53 bell_die -rw------- 1 dfpedro news 1707 Apr 23 15:05 bill -rw-r----- 1 dfpedro news 8416 May 1 06:38 cellular -rw-r----- 1 dfpedro news 942 May 15 12:19 dead.article -rw-r----- 1 dfpedro news 8827 Mar 13 09:38 flog -rw------- 1 dfpedro news 1947 Mar 15 09:30 lane.out $> ls -l | awk '{print $5,$6, $7, $8, $9}' 12956 May 8 22:47 Articles 3626 Mar 13 14:53 bell_die 1707 Apr 23 15:05 bill 8416 May 1 06:38 cellular 942 May 15 12:19 dead.article 8827 Mar 13 09:38 flog 1947 Mar 15 09:30 lane.out There you go. dfpedro@uswnvg.UUCP
jpr@jpradley.jpr.com (Jean-Pierre Radley) (05/21/91)
In article <1991May12.085633.10066@nuchat.sccsi.com> robby@nuchat.sccsi.com (Robert Oliver Jr.) writes: >Hi there, > > Has anybody out there ever printed out the size,date and name of a file >without the permitions and owners just the size,date & name. I would like >to know how you did it! I've never done it, but if I needed to, I'd use sed, cut, awk, or expr. Awk would be the clearest to understand. Jean-Pierre Radley Unix in NYC jpr@jpr.com jpradley!jpr CIS: 72160,1341
antoine@cl.bull.fr (Antoine Haraoui) (05/27/91)
I suggest you try the following : ll | sed -e "s/ */ /g" | cut -d" " -f3,5,6,7 I tried it and it works fine. You can still pipe it to one more sed at the end to format the output the way you like it. Antoine G. HARAOUI, Bull S.A. e-mail: Antoine.Haraoui@cl.bull.fr Software Development Methodology tel : (33-1) 3080 7629 Bulltel->2377629 Av. Jean-Jaures, PC F30D01, fax : (33-1) 3080 7078 Bullfax->2377078 78340 Les Clayes sous Bois, France