strtotime

Hay, i have a database holding events. There are 2 fields 'start' and 'end', these contain timestamps. When an admin enters these dates, they only have the ability to set the day,month,year. So we are only dealing with stamps containing days,months,years, not hours,minutes,seconds (hours,minutes and seconds are set to 0,0,0). I have an event with the start time as 1262304000 and the end time as 1262908800. These convert to Jan 1 2010 and Jan 8 2010. How would i get all the days between these timestamps? I want to be able to return Jan 2 2010 (1262390400), Jan 3 2010 (1262476800) .. all the way

I am not sure why strtotime() in PHP returns different result in different timezone even though same date is given as parameter, does anyone know the answer? I also want to know, can I do similar task (converting a datetime to an int to do calculations easily) with another function which gives same result across different timezone? EDIT: An example: If I use strtotime('2011-09-19 00:00:00') shouldn't it just return the difference between 'January 1 1970 00:00:00' and '2011-09-19 00:00:00' in seconds ? Why timezone is an issue here? And can I get something which gives just difference without

I'm trying to group together dates into a week number and year, and then I want to convert that week number back into a unix timestamp. How can I go about doing this? I assume you are using ISO 8601 week numbers, and want the first day of a ISO 8601 week so that e.g. Week 1 of 2011 returns January 3 2011 . strtotime can do this out of the box using the {YYYY}W{WW} format: echo date("Y-m-d", strtotime("2011W01")); // 2011-01-03 Note that the week number needs to be two digits. Shamefully, DateTime::createFromFormat , the fancy new PHP 5 way of dealing with dates, seems unable to parse this kind

Is there a way to check to see if a date/time is valid you would think these would be easy to check: $date = '0000-00-00'; $time = '00:00:00'; $dateTime = $date . ' ' . $time; if(strtotime($dateTime)) { // why is this valid? } what really gets me is this: echo date('Y-m-d', strtotime($date)); results in: "1999-11-30", huh? i went from 0000-00-00 to 1999-11-30 ??? I know i could do comparison to see if the date is either of those values is equal to the date i have but it isn't a very robust way to check. Is there a good way to check to see if i have a valid date? Anyone have a good function to

1 //php获取前一个小时的时间： 2 3 $mtime= date("Y-m-d H:i:s", strtotime("-1 hour")); 4 //php获取前一天的时间： 5 6 $mtime= date("Y-m-d H:i:s", strtotime("-1 day")); 7 //php获取三天前的时间： 8 9 $mtime= date("Y-m-d H:i:s", strtotime("-3 day")); 10 //php获取前一个月的时间： 11 12 $mtime= date("Y-m-d H:i:s", strtotime("-1 month")); 13 //php获取三个月前的时间： 14 15 $mtime= date("Y-m-d H:i:s", strtotime("-3 month")); 16 //php获取前一年的时间： 17 18 $mtime= date("Y-m-d H:i:s", strtotime("-1 year")); 来源： https://www.cnblogs.com/realredboy/p/10748171.html

获取前一天的时间： $mytime= date("Y-m-d H:i:s", strtotime("-1 day")); 获取三天前的时间： $mytime= date("Y-m-d H:i:s", strtotime("-3 day")); 获取前一个月的时间： $mytime= date("Y-m-d H:i:s", strtotime("-1 month")); 获取前3个月的时间： $mytime= date("Y-m-d H:i:s", strtotime("-3 month")); 获取前一个小时的时间： $mytime= date("Y-m-d H:i:s", strtotime("-1 hour")); 获取前一年的时间： $mytime= date("Y-m-d H:i:s", strtotime("-1 year")); 来源： http://www.cnblogs.com/hgj123/p/4915173.html

获取前一天的时间： $mytime= date("Y-m-d H:i:s", strtotime("-1 day")); 获取三天前的时间： $mytime= date("Y-m-d H:i:s", strtotime("-3 day")); 获取前一个月的时间： $mytime= date("Y-m-d H:i:s", strtotime("-1 month")); 获取前3个月的时间： $mytime= date("Y-m-d H:i:s", strtotime("-3 month")); 获取前一个小时的时间： $mytime= date("Y-m-d H:i:s", strtotime("-1 hour")); 获取前一年的时间： $mytime= date("Y-m-d H:i:s", strtotime("-1 year")); 来源： http://www.cnblogs.com/yxwkf/p/5058655.html