Reference 10th submatch

Question

Normally when you call sed, you can call \1 to reference the submatch number one, i.e.:

<something> | sed "s/<regex>\(<regex>\)<regex>/\1/"

This will just take the middle regex and remove the outer ones.

But what do you do, if you have more then 9 matches? Simply writing \10 doesn't work, because it will be interpreted as take submatch number one and add a zero behind it.

Can you simplify it to reduce the number of capturing groups? — nhahtdh, Aug 06 '12 at 17:46

score 7 · Accepted Answer · edited May 23 '17 at 12:26

7

It's not possible in sed. You'll have to use a different tool.

Here are similar questions on SO:

Circumvent the sed backreference limit \1 through \9.

how to get 10th grouping value in sed?

edited May 23 '17 at 12:26

Community

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answered Aug 06 '12 at 17:56

Andrew Cheong

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thanks, the first one actually says (which I can only test tomorrow) that you can use perl instead of sed and apply a "s/in/out/" expression nearly in the same fashion but much more powerful. That is actually quite, quite helpful! So I voted for duplicate of your first link. – erikbstack Aug 06 '12 at 19:27

score 1 · Answer 2 · answered Aug 06 '12 at 17:42

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I'm not sure about SED, but in many languages (javascript, ruby), that have regex implementations that allow for submatching (or backreferencing as it is sometimes called), only support \1 through \9. \10 is simply not allowed.

Any additional backreferencing would require the use of a more sophisticated parser.

answered Aug 06 '12 at 17:42

Hans Z

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I think Perl allows it, though. – Andrew Cheong Aug 06 '12 at 17:55
Some regex implementations, such as Python, _will_ accept `\10`, etc, _if_ there are more than 9 capture groups in the pattern. – MRAB Aug 06 '12 at 19:12

Reference 10th submatch

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