Escape wildcard problems.

On Fri, Oct 24, 2008 at 08:12:38AM -0700, Gauthier, Dave wrote:

select name from templates where name like '%\_cont\_%';

name
----------------------------------
cgidvcontrol
x8idvcontrol
etc....

I would expect to NOT see these because the "cont" is not preceded by
and followed by an underscore (because I escaped them with \).

You need to escape the escape! backslash is the escape character in
literals as well as like patterns, so you need to double it up. I think
you want to be doing:

name LIKE '%\\_cont\\_%'

Sam

On Friday 24 October 2008, "Gauthier, Dave" <dave.gauthier@intel.com> wrote:

I read in the docs (section 9.7.1) that the backslash... \ ... is the
default escape char to use in "like" expressions. Yet when I try it, it
doesn't seem to work the ay I expect. Here's an example...

select name from templates where name like '%\_cont\_%';

Use double \\ for underscores. I don't know why it's necessary, but it works
here.

--
Alan

Alan Hodgson wrote:

On Friday 24 October 2008, "Gauthier, Dave" <dave.gauthier@intel.com> wrote:

I read in the docs (section 9.7.1) that the backslash... \ ... is the
default escape char to use in "like" expressions. Yet when I try it, it
doesn't seem to work the ay I expect. Here's an example...

select name from templates where name like '%\_cont\_%';

Use double \\ for underscores. I don't know why it's necessary, but it works
here.

Here's why. See the documentation for more information:

craig=> show standard_conforming_strings;
standard_conforming_strings
-----------------------------
off
(1 row)

craig=> SELECT '%\_cont\_%';
WARNING: nonstandard use of escape in a string literal
LINE 1: SELECT '%\_cont\_%';
^
HINT: Use the escape string syntax for escapes, e.g., E'\r\n'.
?column?
----------
%_cont_%
(1 row)

craig=> SELECT E'%\\_cont\\_%';
?column?
------------
%\_cont\_%
(1 row)

craig=> set standard_conforming_strings = 1;
SET

craig=> SELECT '%\_cont\_%';
?column?
------------
%\_cont\_%
(1 row)

--
Craig Ringer

Or you could use:

SELECT name
FROM templates
WHERE name ~ '\_cont\_';

This does it as a regular expression.

~* '\_aa\_';

On Fri, Oct 24, 2008 at 5:07 PM, Craig Ringer
<craig@postnewspapers.com.au> wrote: