pattern by using regular expressions match

I'm playing (and wrong) with regular expressions

Select regexp_substr (1 < PSN > / # < 231 > # < 3 25 3 > / < ABc > ',' < [[: alnum:]] + > ') twice;

give < PSN > - I understand that

Select regexp_substr (1 < PSN > / # < 231 > # < 3 25 3 > / < ABc > ',' < [[: alnum:]] + >, 1.2 ') twice;

give < 231 > - I understand that

Select regexp_substr (1 < PSN > / # < 231 > # < 3 25 3 > / < ABc > ',' < [[: alnum:]] + >, 1.3 ') twice;

give < ABc > which confused me until I realized that < 3 25 > has not been matched, because it has a space inside

so I changed it to

Select regexp_substr (1 < PSN > / # < 231 > # < 3 25 3 > / < ABc > ',' <. * + > ', 1.3) twice;

who gave a syntax error, so I changed it again to

Select regexp_substr (1 < PSN > / # < 231 > # < 3 25 3 > / < ABc > ',' <. * + > ') twice;

that works, but gives < PSN > / # < 231 > # < 3 25 3 > / < ABc >

I guess because the. * corresponds to anyting that all but the last closure >

The question I have is how can I retrieve the text of mounting bracket included (even if it has multiple spaces)?

Hello

9c5dfde3-EAAE-45A7-80a1-bba8a71c826c wrote:

Thanks for that - is there a way to return all 4 surrounded by extracts of <> without resorting to PL/SQL?

Of course;

REGEXP_REPLACE (str

, '(^|>)[^<>

, '\1'

)

Returns a copy of str with all outside rafters removed, for example

<231><3 25="">

.  It doesn't matter how many pairs of sharp hooks - there is.

Tags: Database

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