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I need to extract the public key in the shell script, Not sure how to use this. I am new to shell script. https://regex101.com/r/SXDEaU/1

Content of the file:

public_key=#STARTKEY#<public key base64 encoded>#ENDKEY#

Regex: /public_key=#STARTKEY#(.*)#ENDKEY#/s

Since the key is base64 encoded, it is a multiline string.

Desired output: <public key base64 encoded>

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    Please add your desired output (no description, no images, no links) for that sample input to your question (no comment). Commented Dec 9, 2021 at 6:09
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    grep -Pzo '(?<=public_key=#STARTKEY#)[\s\S]*(?=#ENDKEY#)' file ? Commented Dec 9, 2021 at 6:12

2 Answers 2

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With awk using its RS variable and setting it to paragraph mode please try following code.

awk -v RS= 'match($0,/public_key=#STARTKEY#.*#ENDKEY#/){print substr($0,RSTART+21,RLENGTH-29)}'  Input_file
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That doesn't support blanks lines. But while allowed, one is extremely unlikely to encounter any. So this is probably not a problem.
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perl -M5.010 -0777ne'say for /public_key=#STARTKEY#(.*?)#ENDKEY#/s' file

The key is -0777 which causes the entire file to be read in as one line.

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I would use ([^#]*) instead of (.*), then strip all non base64 characters (like newlines) from the match
@Fravadona, I didn't look at the pattern; I just showed how to use it as requested. [^#]* isn't correct, That should be (?:(?!#ENDKEY#).)*. But .*? would do just as well here, and it would be better than .*. Adjusted answer. /// Stripping non-base64 characters wasn't requested, isn't always desirable, and can't be done without knowing which flavour of base64 is being used.
Oh yes, of course perl has non greedy *

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