On Sun, 25 Apr 2010 21:39:43 +0100, Alberto Simões <hashas...@gmail.com> wrote:
>Hello > > I am running on the fly a query to count the number of > words starting with one of the 26 letters. > > I am doing the usual SELECT COUNT(term) from dictionary WHERE normword > LIKE "a%" (for the 26 letters) > >normword is the term normalized without accents and the like Would your application allow to return all 26 in one query? SELECT COUNT(term) FROM dictionary WHERE normword >= 'a' AND normword <= 'zzzzzzzzzz' GROUP BY substr(normword,1,1); (untested, but certainly faster than 26 separate queries) >Is there any way to make this query faster? It is taking about 10 >second for 140K entries. > >One idea is to add a column named 'letter' and SELECT COUNT(letter) >from dictionary WHERE letter = 'a'. >But are there other solutions? > >Thanks -- ( Kees Nuyt ) c[_] _______________________________________________ sqlite-users mailing list sqlite-users@sqlite.org http://sqlite.org:8080/cgi-bin/mailman/listinfo/sqlite-users
