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

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

>Solution :

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.