Re: using date as an "id" field...

try this one date("Y-m-d",time()-24*3600); -- Hidulf http://www.hidulf.com "Sc" <sven@gstar.com.au> wrote in message news:20011009024925.7474.qmail@pb1.pair.com... > Thanks for the response but it doesn't seem to work for me... > This is todays date reference: 2001-10-09 > And this is meant to be the previous entry in the sql database: > 1999-11-29 > > > In article <20011008101141.59851.qmail@pb1.pair.com>, "Piotr Martyniak" > <martynia@box43.gnet.pl> wrote: > > > use: > > ereg("([0-9]{4})-([0-9]{1,2})-([0-9]{1,2})", $your_data, $numbers); > > $day_before = mktime(0, 0, 0, $numbers[2], $numbers[3], $numbers[1]) - > > 24 > > * 3600; > > $new_date = date("Y/n/j", $day_before); > > should work natur > > > >