Timeline for How can I output only captured groups with sed?

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May 15, 2017 at 19:36	comment	added	abalter		Also, you can't do `echo "a 10 b 12" \| grep -Eo "a ([0-9]+)"` and get just the "10". But this works: `echo "a 10 b 12" \| grep -Eo "a ([0-9]+)" \| sed 's/a //'`
Oct 3, 2012 at 17:56	comment	added	idbrii		@Pablo: grep's only outputting what matches. To give it multiple groups, use multiple expressions: `grep -Eow -e "[0-9]+" -e "[abc]{2,3}"` I don't know how you could require those two expressions to be on one line aside from piping from a previous grep (which could still not work if either pattern matches more than once on a line).
Mar 14, 2011 at 8:30	comment	added	doc_id		Hello Michael. Did you managed to extract nth captured group by grep ?
May 6, 2010 at 12:11	comment	added	Pablo		@Bert F: I understand the matching part, but it's not capturing group. What I want is to have like this ([0-9]+).+([abc]{2,3}) so there are 2 capturing groups. I want to output ONLY capturing groups by backreferences or somehow else.
May 6, 2010 at 11:36	comment	added	Bert F		@Michael - that's why the `o` option is there - unixhelp.ed.ac.uk/CGI/man-cgi?grep : -o, --only-matching Show only the part of a matching line that matches PATTERN
May 6, 2010 at 1:24	comment	added	Pablo		@ghostdog74: Absolutely agree with you. How can I get greo to output only captured groups?
May 6, 2010 at 1:11	history	answered	ghostdog74	CC BY-SA 2.5