Extract value between double quotes

Question

My query is to extract the value between double quotes "". Sample input is:

10.219.41.68 - - - [11/Jun/2014:10:23:04 -0400] Sec:0 MicSec:1797 "GET /balancer-manager HTTP/1.1" 200 28980 "-" "curl/7.15.5 (i386-redhat-linux-gnu) libcurl/7.15.5 OpenSSL/0.9.8b zlib/1.2.3 libidn/0.6.5"

I have large log files, so values can be vary for each line, need to extract the value between first occurrence of double quotes…

Expected output:

GET /balancer-manager HTTP/1.1

Anyone have any idea then please suggest.

https://stackoverflow.com/questions/7804673/escaping-separator-within-double-quotes-in-awk — Ciro Santilli OurBigBook.com, Aug 05 '17 at 09:02

score 37 · Answer 1 · answered Aug 04 '14 at 10:05

37

You can just use cut for this:

$cut -d '"' -f2 < logfile
GET /balancer-manager HTTP/1.1

-d '"' tells cut to use a double quote as its field delimiter. -f2 tells it to take the second field, which is between the first and second quotes - or the first quoted string, exactly what you want.

answered Aug 04 '14 at 10:05

Michael Homer

76,565

How to print all even numbered columns in cut easily like %!cut -d '"' -f2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50,52,54,56,58? – hhh Jul 20 '17 at 18:46

score 13 · Answer 2 · edited May 01 '17 at 00:06

13

A way using awk

awk -F'"' '$0=$2' file

If for some absurd reason your HTTP methods are actually 0 and you want to output these

awk -F'"' '{$0=$2}1' file

edited May 01 '17 at 00:06

phk

5,953
7
42
71

answered Aug 04 '14 at 10:14

What if $2 is null? – cuonglm Aug 04 '14 at 10:19
Then it wont print it ? The same as it wont in your answer. Only difference is your answer will print a blank line for every record with no $2. – Aug 04 '14 at 10:21
No, my answer print a blank like. And imagine that the value is 0, not only null? – cuonglm Aug 04 '14 at 10:23
I'd imagine given the context of the question that wouldn't be a problem. – Aug 04 '14 at 10:24
Thanks all for your quick responses...Its really helpful for me. Thanks a lot! – user79658 Aug 05 '14 at 10:18

score 4 · Answer 3 · answered Aug 04 '14 at 10:12

4

You can do it many ways.

With awk:

$ awk -F'"' '{print $2}' file
GET /balancer-manager HTTP/1.1

With perl:

$ perl -F'"' -anle 'print $F[1]' file
GET /balancer-manager HTTP/1.1

answered Aug 04 '14 at 10:12

cuonglm

153,898

score 4 · Answer 4 · edited May 01 '17 at 00:06

4

Since an awk and perl solutions are already provided, I wanted to try sed:

sed 's/[^"]*"\([^"]*\)".*/\1/' file

edited May 01 '17 at 00:06

phk

5,953
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answered Aug 06 '14 at 09:05

Srini

161

score 1 · Answer 5 · answered Jun 12 '17 at 10:30

1

processing quoted input numbers

echo   1234   | awk '{                i=strtonum($1) ;  printf( "%s %d\n",$1, i)}'  # no problem
echo '"1234"' | awk '{                i=strtonum($1) ;  printf( "%s %d\n",$1, i)}'  # does not work
echo '"1234"' | awk '{ gsub("\"",""); i = $1         ;  printf( "%s %d\n",$1, i)}'  # works

answered Jun 12 '17 at 10:30

A-Man

11

1

Try to be a little bit more verbose in your answer – ddnomad Jun 12 '17 at 10:35
the only answer to mention awk's gsub; worked for my use case – jwd630 Apr 15 '20 at 20:58

Extract value between double quotes

5 Answers5

processing quoted input numbers