Re: BUG #6057: regexp_replace & back references
От | Alvaro Herrera |
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Тема | Re: BUG #6057: regexp_replace & back references |
Дата | |
Msg-id | 1308080704-sup-5009@alvh.no-ip.org обсуждение исходный текст |
Ответ на | BUG #6057: regexp_replace & back references ("Marc Mamin" <marc@intershop.de>) |
Список | pgsql-bugs |
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. -- Ãlvaro Herrera <alvherre@commandprompt.com> The PostgreSQL Company - Command Prompt, Inc. PostgreSQL Replication, Consulting, Custom Development, 24x7 support
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