Get JSON value using grep, sed, or awk

Question

I have a JSON configuration file I'm trying to extract values from via the following grep command:

grep -oP '"private_key": "\K[^\"]*' config.json

As you can see, I have working regex, however the \K syntax is not supported in older versions of Regex. For example, if you switch the flavor to Python or ECMAscript, the regex no longer works. I'm wondering if there's some other syntax to achieve the same result.

https://regex101.com/r/jfGwNT/1

Also, if anyone can think of a better name for this question, please let me know!

Will you consider other tools like sed/awk? – anubhava Apr 08 '22 at 17:14 — anubhava, Apr 08 '22 at 17:14
(I think your title is fine) – Poul Bak Apr 08 '22 at 17:16 — Poul Bak, Apr 08 '22 at 17:16
I would use sed or awk, yes – Brandon McAlees Apr 08 '22 at 17:19 — Brandon McAlees, Apr 08 '22 at 17:19

anubhava · Accepted Answer · 2022-04-08T17:50:11.350

3

Instead of gnu-grep, you can use sed like this:

sed -nE 's/^ *"private_key": "([^"]+)".*/\1/p' file.json

-----BEGIN PRIVATE KEY-----\nMyKey\n-----END PRIVATE KEY-----\n

edited Apr 08 '22 at 17:50

answered Apr 08 '22 at 17:20

anubhava

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This is awesome, but how would I remove the comma from the end? I just want what is between the double quotes. – Brandon McAlees Apr 08 '22 at 17:26
Yes that can be done. See my edited answer. – anubhava Apr 08 '22 at 17:50

Get JSON value using grep, sed, or awk

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