1

MY File:

"DocumentCreationDate="2019-07-15T23:56:31" SampleID="1" entClassID="65535" ClientID="0" CardID="11209797""

Want to grep Pattern :

CardID="11209797"

The number may be different between " "

Command used : egrep -o CardID='\"[^]"*]'

OUTPUT : CardID="1

Desired Outpt: CardID="11209797"

2 Answers 2

2

Try this,

grep -o 'CardID="[^"]*"' file
CardID="11209797"
1
$ grep -Eo 'CardID="[^"]*"' file
CardID="11209797"

Notes:

  1. egrep is deprecated. Use grep -E instead.

  2. Inside single-quotes, " does not need to be escaped.

  3. [^]"*] matches any one character that is not ], ", or *. What you want instead is to any series of characters except ". To do that, use [^"]*.

2
  • Your regexp is a BRE, no need for -E to enable EREs.
    – Ed Morton
    Jul 16, 2019 at 13:09
  • 1
    worked , resolved Jul 17, 2019 at 3:09

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