Date: Wed, 26 Nov 2025 15:48:44 +0100
From: Vincent Veyron <vv.lists@wanadoo.fr>
To: pgsql-general@lists.postgresql.org
Subject: Re: Selecting all variations of job title in a list
Message-Id: <20251126154844.e6e3d5c7f476cc966fc201f4@wanadoo.fr>
In-Reply-To: <c5296761-2f10-a59b-ad16-a0189e57cd5a@appl-ecosys.com>
References: <c5296761-2f10-a59b-ad16-a0189e57cd5a@appl-ecosys.com>
Mime-Version: 1.0
Content-Type: text/plain; charset=ISO-8859-1
Content-Transfer-Encoding: quoted-printable
Archived-At: <https://www.postgresql.org/message-id/20251126154844.e6e3d5c7f476cc966fc201f4%40wanadoo.fr>
Precedence: bulk

On Tue, 25 Nov 2025 08:33:07 -0800 (PST)
Rich Shepard <rshepard@appl-ecosys.com> wrote:

> Companies can have slightly different titles for the same job; for example
> (using abbreviations). 'Asst Gen Mgr.', 'Env Mgr,', 'Gen Mgr,'. 'Mgr,',
> 'Plant Mgr.'
>=20
> I want to select all people table rows that contain these varieties. I kn=
ow
> the 'like' operator uses '%' as a wildcard, but is not accepted in an 'in'
> list.
>=20
> Is there a way to use a multicharacter wildcard in an 'in' list?
>=20

Hi Rich,

As Ron wrote, a regexp would do it, I guess?

SELECT * FROM people WHERE title ~ 'Asst Gen Mgr|Env Mgr|Gen Mgr|Mgr|Plant =
Mgr';

Although in your example, you would get the same result with just:

SELECT * FROM people WHERE title ~ 'Mgr';

https://www.postgresql.org/docs/current/functions-matching.html#FUNCTIONS-P=
OSIX-REGEXP

--=20

					Bien =E0 vous, Vincent Veyron=20

https://marica.fr/=20
Logiciel de suivi des contentieux juridiques, des sinistres d'assurance et =
des contrats