RE: [PHP-DB] Count database-values

> -----Original Message----- > From: Becoming Digital [mailto:subs@becomingdigital.com] > Sent: 10 June 2003 12:15 > > I screwed up my own code. Silly me. It should read: Surely your first attempt is the right one? There's only ever going to be 1 Total, so why waste a while loop trying to read more than one result row? > $query = "SELECT SUM(bedrag) AS Total FROM finance WHERE > posneg = 'af'"; > $result = mysql_query( $query ); > while ( $total = mysql_fetch_array( $result ) ) > { > $total = $total['Total']; > } > > Edward Dudlik > Becoming Digital > www.becomingdigital.com > > > ----- Original Message ----- > From: "Becoming Digital" <subs@becomingdigital.com> > To: <php-db@lists.php.net> > Sent: Tuesday, 10 June, 2003 06:43 > Subject: Re: [PHP-DB] Count database-values > > > > So far no problems, but I want to count all these values. > > I tried with SUM but with the following code it doesn't work: > > > > $squery = "SELECT SUM(bedrag) AS Total FROM finance WHERE > posneg = 'af'"; > > > > any suggestions how to do? I want to store it in a > variabele ($total) > > so I can echo it. (E.g.: $total = 93.63 in this case) > > If you want the total, you need to do something like the following: > > $query = "SELECT SUM(bedrag) AS Total FROM finance WHERE > posneg = 'af'"; > $result = mysql_query( $query ); > $total = mysql_fetch_array( $result ); > $total = $total['Total']; > Cheers! Mike --------------------------------------------------------------------- Mike Ford, Electronic Information Services Adviser, Learning Support Services, Learning & Information Services, JG125, James Graham Building, Leeds Metropolitan University, Beckett Park, LEEDS, LS6 3QS, United Kingdom Email: m.ford@lmu.ac.uk Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211