如何从命令输出中获取第二列？

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 double-quoted string.

My purpose is to get the second column only, like:

"A B"
"C"
"D"

I intended to use <some_command> | awk '{print $2}' <some_command> | awk '{print $2}' to accomplish this. But the question is, some values in the second column contain space(s), which happens to be the default delimiter for awk to separate the fields. Therefore, the output is messed up:

"A
"C"
"D"

How do I get the second column's value (with paired quotes) cleanly?

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

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 3

This should work to get a specific column out of the command output "docker images":

REPOSITORY                          TAG                 IMAGE ID            CREATED             SIZE
ubuntu                              16.04               12543ced0f6f        10 months ago       122 MB
ubuntu                              latest              12543ced0f6f        10 months ago       122 MB
selenium/standalone-firefox-debug   2.53.0              9f3bab6e046f        12 months ago       613 MB
selenium/node-firefox-debug         2.53.0              d82f2ab74db7        12 months ago       613 MB


docker images | awk '{print $3}'

IMAGE
12543ced0f6f
12543ced0f6f
9f3bab6e046f
d82f2ab74db7

This is going to print the third column

Answer 4

或者使用sed和regex。

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

Answer 5

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

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

or:

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

Answer 6

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

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

Answer 7

awk -F"|" '{gsub(/\"/,"|");print "\""$2"\""}' your_file

Answer 8

#!/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