How to get the second column from command output?

Question

My command\'s output is something like:

1540 \"A B\"
   6 \"C\"
 119 \"D\"

The first column is always a number, followed by a space, then a

Answer 1

Use -F [field separator] to split the lines on "s:

awk -F '"' '{print $2}' your_input_file

or for input from pipe

<some_command> | awk -F '"' '{print $2}'

output:

A B
C
D

Answer 2

#!/usr/bin/python
import sys 

col = int(sys.argv[1]) - 1

for line in sys.stdin:
    columns = line.split()

    try:
        print(columns[col])
    except IndexError:
        # ignore
        pass

Then, supposing you name the script as co, say, do something like this to get the sizes of files (the example assumes you're using Linux, but the script itself is OS-independent) :-

ls -lh | co 5

Answer 3

If you could use something other than 'awk' , then try this instead

echo '1540 "A B"' | cut -d' ' -f2-

-d is a delimiter, -f is the field to cut and with -f2- we intend to cut the 2nd field until end.

Answer 4

You don't need awk for that. Using read in Bash shell should be enough, e.g.

some_command | while read c1 c2; do echo $c2; done

or:

while read c1 c2; do echo $c2; done < in.txt

Answer 5

If you have GNU awk this is the solution you want:

$ awk '{print $1}' FPAT='"[^"]+"' file
"A B"
"C"
"D"

Answer 6

Or use sed & regex.

<some_command> | sed 's/^.* \(".*"$\)/\1/'