BUG #6057: regexp_replace & back references
The following bug has been logged online:
Bug reference: 6057
Logged by: Marc Mamin
Email address: marc@intershop.de
PostgreSQL version: 9.0.4
Operating system: Windows
Description: regexp_replace & back references
Details:
select regexp_replace ('a','(a)','\\1'||substring('\\1',1,1)||'\\1','g')
= a\1
I'd expect a\1a as result.
More generally, I miss the avaibility to use back references as function
parameters. It is only possible with some operators (e.g. '\\1'||'\\1'),
probably depending on their lexical precedence. Any way to workaround this
limitation would be very helpful.
HTH,
Marc Mamin
Excerpts from Marc Mamin's message of mar jun 14 12:31:34 -0400 2011:
The following bug has been logged online:
Bug reference: 6057
Logged by: Marc Mamin
Email address: marc@intershop.de
PostgreSQL version: 9.0.4
Operating system: Windows
Description: regexp_replace & back references
Details:select regexp_replace ('a','(a)','\\1'||substring('\\1',1,1)||'\\1','g')
= a\1
I'd expect a\1a as result.
Note that the substring returns a single character which is a literal \.
That \ escapes the \ in the final '\\1', which turns into the literal \
that you see in the result. The subsequent 1 is the leftover char from
the final '\\1'.
I think this is clearer if you set standard_conforming_strings to on.
--
Álvaro Herrera <alvherre@commandprompt.com>
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