272

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}' 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?

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  • 1
    stackoverflow.com/questions/2961635/… Commented Apr 21, 2013 at 22:39
  • 1
    I tried using awk '{$1=""; print $0}', but it still has a leading white space character. It could be removed by sed '/^ //'. Yet, could this be done with awk? Commented Apr 21, 2013 at 23:00

8 Answers 8

Reset to default
325

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
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5 Comments

This is good, but I also want the original surrounding quotes. Could it be done? Thanks.
you could cheat, and change awk's print to '{print "\""$2"\""}'
Yup, this works. Thanks a lot, Alex! By the way, so many quotes, :)
@Alex, could you explain how you used the double quotation marks and backslash to get what the op wanted.
@Timo The quotes and backslashes breakdown can be envisioned as "\"" + $2 + "\"". The surrounding quotation marks are indicating something to be appended to the output, and the escaped quotation mark (\") is being printed. To help visualize it, this is what it would look like if we wanted to add blank spaces around $2 instead of quotation marks: '{print " "$2" "}'. We can also add format spacing to make it a little easier to grok: '{print " " $2 " "}'
131

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.

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5 Comments

this helped me trying to do following (fetch commit id of a file in git): git annotate myfile.cpp | grep '2016-07' | head -1| cut -f1
This is good, but does not work if delimiter is more than one character long. That's where the awk solution comes in handy
Why is a space not used after -d? It looks a bit odd in that way.
if your output has more than one column and you only need the second column, use cut -d' ' -f2
@ChrisStryczynski: You can also do: ` cut -d\ -f2-` (note: two spaces after the back-slash!) Does that look less odd ?
108

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

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5 Comments

Have you seen out put of docker images | awk '{print $5}' ?
@ShashiRanjan No, what was it?
This seems to break on any row with an empty value.
This breaks as "IMAGE ID" has a space in between.
Why does this work despite not specifiying the field separator?
37

Or use sed & regex.

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

Shorter cmd as you do not need start and end markers: <some_command> | sed 's/.* \(".*"\)/\1/'
It would be nice if anyone could explain a little bit on what 's/^.* \(".*"$\)/\1/' stands for
^ marks the beginning of the line .* means any sequence of characters \( and \) marks a « captured » group of character, what we will treat as \1 in the Right hand side of the sed sequence ".*" means any sequence of chars between quotes $ marks the end of the line
25

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
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2 Comments

You should always use the -r argument with read, especially when you don't know what the input will be. Otherwise backslashes will mess things up.
If using this beware that if the "read" column count is less than the input columns then the last variable will contain all the remaining fields (ie. the last variable might not contain only that column).
13

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

$ awk '{print $1}' FPAT='"[^"]+"' file "A B" "C" "D"
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1 
awk -F"|" '{gsub(/\"/,"|");print "\""$2"\""}' your_file
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0 
#!/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

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