Thread: 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.

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