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ls -l (without showing total blocks)

 
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# 1  
Old 12-07-2009
ls -l (without showing total blocks)
Is it possible to use the ls command and NOT see the total blocks? For example, instead of seeing this...

Code:
$ ls -l
total 16
drwxr-xr-x 2 voicem voicem 4096 Dec  3 09:19 directory1
drwxr-xr-x 2 voicem voicem 4096 Dec  1 01:29 directory2
drwxr-xr-x 2 voicem voicem 4096 Dec  2 00:47 directory3
drwxr-xr-x 2 voicem voicem 4096 Nov 30 00:10 directory4

I'd like to see this...

Code:
$ ls -l
drwxr-xr-x 2 voicem voicem 4096 Dec  3 09:19 directory1
drwxr-xr-x 2 voicem voicem 4096 Dec  1 01:29 directory2
drwxr-xr-x 2 voicem voicem 4096 Dec  2 00:47 directory3
drwxr-xr-x 2 voicem voicem 4096 Nov 30 00:10 directory4

Notice that the desired output does not list total 16. Is this doabale?
# 2  
Old 12-07-2009
ls -l | sed /^total/d
# 3  
Old 12-07-2009
or even
Code:
ls -l|sed '/^t/d'

in this case Smilie
-or-
Code:
ls -l|grep -v ^t

-or
Code:
ls -l|grep ^[^t]

-or-
Code:
ls -l|awk 'NF>2'

-or-
Code:
ls -l|awk 'NR>1'

-or
Code:
ls -l|awk 'NF-2'

-or
Code:
ls -l|awk 'NR-1'

-or
Code:
ls -l|awk p++

-or-
Code:
ls -l|sed 1d

Last edited by Scrutinizer; 12-07-2009 at 06:23 PM..
# 4  
Old 02-23-2010
MySQL
see the following example:
it also prints from second line from last line.
Code:
ls -l | sed  -n '2,$p'

# 5  
Old 02-23-2010
How about this ?

Code:
ls -l | tail -n +2

# 6  
Old 02-23-2010
ls -l output
Try this also,

ls -l | sed -n '1!P'
 
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