How do i get the value present in first double quotes? - bash

Question

I'm currently writing a bash script to get the first value among the many comma separated strings.
I have a file that looks like this -
name
things: "water bottle","40","new phone cover",10
place
I just need to return the value in first double quotes.
water bottle
The value in first double quotes can be one word/two words. That is, water bottle can be sometimes replaced with pen.
I tried -
awk '/:/ {print $2}'
But this just gives
water
I wanted to comma separate it, but there's colon(:) after things. So, I'm not sure how to separate it.
How do i get the value present in first double quotes?
EDIT:
SOLUTION:
I used the below code since I particularly wanted to use awk -
awk '/:/' test.txt | cut -d\" -f2

Answer 1

A solution using the cut utility could be
cut -d\" -f2 infile > outfile

Answer 2

Using gnu awk you could make use of a capture group, and use a negated character class to not cross the , as that is the field delimiter.
awk 'match($0, /^[^",:]*:[^",]*"([^"]*)"/, a) {print a[1]}' file
Output
water bottle
The pattern matches
^ Start of string
[^",:]*:Optionally match any value except " and , and :, then match :
[^",]* Optionally match any value except " and ,
"([^"]*)" Capture in group 1 the value between double quotes
If the value is always between double quotes, a short option to get the desired result could be setting the field separator to " and check if group 1 contains a colon, although technically you can also get water bottle if there is only a leading double quote and not closing one.
awk -F'"' '$1 ~ /:/ {print $2}' file

Answer 3

With your shown samples, please try following awk code.
awk '/^things:/ && match($0,/"[^"]*/){print substr($0,RSTART+1,RLENGTH-1)}' Input_file
Explanation: In awk program checking if line starts with things: AND using match function to match everything between 1st and 2nd " and printing them accordingly.

Answer 4

Solution 1: awk
You can use a single awk command:
awk -F\" 'index($1, ":"){print $2}' test.txt > outfile
See the online demo.
The -F\" sets the field separator to a " char, index($1, ":") condition makes sure Field 1 contains a : char (no regex needed) and then {print $2} prints the second field value.
Solution 2: awk + cut
You can use awk + cut:
awk '/:/' test.txt | cut -d\" -f2 > outfile
With awk '/:/' test.txt, you will extract line(s) containing : char, and then the piped cut -d\" -f2 command will split the string with " as a separator and return the second item. See the online demo.
Solution 3: sed
Alternatively, you can use sed:
sed -n 's/^[^"]*"\([^"]*\)".*/\1/p' file > outfile
See the online demo:
#!/bin/bash
s='name
things: "water bottle","40","new phone cover",10
place'
sed -n 's/^[^"]*"\([^"]*\)".*/\1/p' <<< "$s"
# => water bottle
The command means
-n - the option suppresses the default line output
^[^"]*"\([^"]*\)".* - a POSIX BRE regex pattern that matches
^ - start of string
[^"]* - zero or more chars other than "
" - a " char
\([^"]*\) - Group 1 (\1 refers to this value): any zero or more chars other than "
".* - a " char and the rest of the string.
\1 replaces the match with Group 1 value
p - only prints the result of a successful substitution.