linux bash命令由空格分隔

Question

so I'm trying to display only columns at a time

first ls -l gives me this

drwxr-xr-x 11 stuff stuff      4096 2009-08-22 06:45 lyx-1.6.4
-rw-r--r--  1 stuff stuff  14403778 2009-10-26 02:37 lyx.tar.gz

I'm using this:

ls -l |cut -d " " -f 1 

to get this

drwxr-xr-x 
-rw-r--r-- 

and it displays my first column just fine. Then I want to see on the second column

ls -l |cut -d " " -f 2

I only get this

11

Shouldn't I get

11
1

?
Why is it doing this?

if I try

   ls -l |cut -d " " -f 2-3

I get

11 stuff

There's gotta be an easier way to display columns right?

Answer 1

这应该显示第二列：

ls -l | awk '{print $2}'

Answer 2

cut considers two sequential delimiters to have an empty field in between. So the second line:

-rw-r--r--  1 stuff stuff

has fields:

1: -rw-r--r--
2: --empty field--
3: 1
etc.

You can use use column fields in cut:

ls -l | cut -c13-14

Or you can use awk to separate fields (unlink, cut awk will treat sequential delimiters as a single delimiter).