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vehomzzz
vehomzzz

Reputation: 44618

Using bash (bash awk, sed) read a file and extract word matching prefix between quotes

I have a file containing (one per line). I would like to extract word between quotes starting with the some pattern. (in my case it is C_)

"PATTERNabcde"  sdfds  sdfds
"sdfsdfsdf"   sdfdsf sdfdsf
" PATTERNabc"          dfdsdfd

and I want to extract: PATTERNabcde PATTERNabc

EDIT:
I would like to ALSO extract word between quotes that don't start with the $PATTERN.

Upvotes: 0

Views: 4724

Answers (3)

marco
marco

Reputation: 4675

You can use awk:

awk -F\" '$2~/^[[:space:]]*PATTERN/{print $2}' file

This works if there is only one word enclosed in quotation marks or if the one that may start with pattern is always the first one; otherwise, you'd have to use a for cycle:

awk -F\" '{for (i=2;i<=NF;i+=2) if ($i ~/^[[:space:]]*PATTERN/ ) {print $i;next}}'

Upvotes: 2

codaddict
codaddict

Reputation: 455342

You can use grep as:

grep -Po '(?<=")\s*PATTERN[^"]*(?=")' file

See it

Upvotes: 2

Matthew Flaschen
Matthew Flaschen

Reputation: 284927

sed -rn 's/.*?".*?(PATTERN[^"]*)".*/\1/p'
  • -r - extended regex
  • -n - disabled auto-print
  • .*? - zero or more characters, non-greedy
  • ( - open capturing group
  • [^"] - any character but "
  • ) - close capturing group
  • \1 - first matching group
  • p - print

We just replace every line with the first group. If there is a replacement, we print.

Upvotes: 2

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