Re: how to use the function "strtotime"

From: Date: Thu, 20 Sep 2001 09:49:02 +0000
Subject: Re: how to use the function "strtotime"
References: 1  Groups: php.general 
Request: Send a blank email to php-general+get-67963@lists.php.net to get a copy of this message
<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 -- ««««««««««««««««««««««««

« previous php.general (#67963) next »