CODEWITHSUNDEEP
bashshellawksed
Sourabrt
Sourabrt

Reputation: 2336

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

Upvotes: 5

Views: 1035

Answers (4)

RavinderSingh13
RavinderSingh13

Reputation: 133538

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.

Upvotes: 4

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626920

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.

Upvotes: 1

The fourth bird
The fourth bird

Reputation: 163362

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

Upvotes: 4

M. Nejat Aydin
M. Nejat Aydin

Reputation: 10123

A solution using the cut utility could be

cut -d\" -f2 infile > outfile

Upvotes: 4

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