Re: how to use the function "strtotime"

thank a lot , I learn a lot from reply from you ! "* R&Ze:" <renze@datalink.nl> wrote in message news:20010920114902.Y17452@pro-pain.telemediair.nl... > <Original message> > From: mydata <muyuan11@sian.com> > Date: Thu, Sep 20, 2001 at 11:41:09AM +0800 > Message-ID: <20010920033656.96921.qmail@pb1.pair.com> > Subject: [PHP] how to use the function "strtotime" > > > Hi, > > I'm php developer beginner in China,but I am confused with function > > "strtotime". > > yes ,I know the usage of function "strtotime" in php manual. But description > > of php manual about it is too little.I want to know more about it. > > </Original message> > > <Reply> > > It's not all that difficult, really. It's exactly how it's described > in the manual. Here is some code that runs some samples: > > (I must say, though, that most examples can be done a lot easier > without strtotime(), but 'cause these are examples of strtotime()... > well... you know :) > > --- PHP Example Code --- > <PRE> > <?php > > /* Show a given date (+time) */ > print ("Date: \n\t"); > print (date ("r", strtotime ("January 31 2000 > 19:20:15"))."\n\n"); > > /* Show the current date (+time) */ > print ("Now: \n\t"); > print (date ("r", strtotime ("now"))."\n\n"); > > /* Show the date (+time) of tomorrow */ > print ("Tomorrow: \n\t"); > print (date ("r", strtotime ("tomorrow"))."\n\n"); > > /* Show the date (+time) of yesterday */ > print ("Yesterday: \n\t"); > print (date ("r", strtotime ("yesterday"))."\n\n"); > > /* Show the date (+time) of yesterday, where now is tomorrow. So the > the output will be today :) */ > print ("Yesterday, where 'now' is tomorrow: \n\t"); > print (date ("r", strtotime ("yesterday", > strtotime("tomorrow")))."\n\n"); > > /* Show the date (+time) 13 days from december 24th 1998 */ > print ("13 days from december 24th 1998: \n\t"); > print (date ("r", strtotime ("+13 days", strtotime("December 24 1998")))."\n\n"» > > /* Let's say you get some date from the db (or something) and you > * want to > add 2 years and 4 months and 23 days. */ > print ("Well... just see: \n\t"); > $date_from_db = "2001-09-20"; > print (date ("r", strtotime ("+2 years +4 months +23 days", > mktime(0,0,0,substr($date_from_db, 5, 2), > substr($date_from_db, 8, 2), > substr($date_from_db, 0, 4))))."\n\n"); > > /* Etc. */ > print ("And so on... :)<HR noshade size=\"1\">"); > > ?> > </PRE> > --- End of PHP Example Code --- > > </Reply> > > -- > > * R&zE: > > > -- »»»»»»»»»»»»»»»»»»»»»»»» > -- Renze Munnik > -- DataLink BV > -- > -- E: renze@datalink.nl > -- W: +31 23 5326162 > -- F: +31 23 5322144 > -- M: +31 6 21811143 > -- > -- Stationsplein 82 > -- 2011 LM HAARLEM > -- Netherlands > -- > -- http://www.datalink.nl > -- ««««««««««««««««««««««««