karl@forest.gsfc.nasa.gov (Karl Anderson) (05/14/91)
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! Do you have the 'cut' command? Here's an alias I have defined in my .cshrc on my Suns: alias sdir 'ls -lta \!* | cut -c33- | more' The "-c33-" option to cut says "print everything from column 33 on." Karl A. Anderson | Internet: karl@forest.gsfc.nasa.gov NASA/GSFC code 923 (STX) | voice: (301) 286-3815 Greenbelt, MD 20771 | #include "std_disclaimer"
GEBORSA%PLKRCY11@searn.sunet.se ( Mariusz Borsa) (05/15/91)
I think that there is another (simpler) solution, which follows
ls -l awk '{print $3 $4 $5}'
Yours
Mariusz Borsaasqe-x-w-ic03@worms-emh1.army.mil (Udo Walter, SYS ADM, CDOIM) (05/15/91)
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 permissions and owners just the size, date & name. I would like >to know how you did it! Try this one. It gives tells you also whether it's a file or a directory. Put it in a scriptfile and make it executable. ls -al $1 | nawk '{if (match($1,"d")) printf("<DIR> %7d %3s %2s %5s %s\n",$5,$6,$7,$8,$9) else printf(" %7d %3s %2s %5s %s\n",$5,$6,$7,$8,$9)}'| more Udo Walter