Get days left

thoand

I pull out two dates from the database, startdate and enddate
how can I get how many days there are between.

eg.

[startdate] - [endate]

2003-03-13
- 2003-03-14

should return one day.

both are datetime fields

regards thomas

epp_b

You'll have to create a variable that will represent the number of days, as well as variables for each unit of time (year, month, and day).

For the years, subtract the endyear from the startyear , and increment that answer by 365 or 366 (depending on whether or not it is a leap year) using a loop and iterates until it equals startyear - endyear. (you can't simply multiply it times 365, because of leap years). A year is a leap year if it is divisible by four, unless it is also divisible by 100. You'll have to check for this through each iteration of the loop.

Then, for the months, you'll have to check which month it is to see how many days it has.

Months with 31 days:
-January (1)
-March (3)
-May (5)
-July (7)
-August (8)
-October (10)
-December (12)

Months with 30 days:
-April
-June
-September
-November

And, of course, you can have a variable be created when it is a leap year that is "true", to check whether February has 28 or 29 days.

The days are the easiest: they're exactly what they say they are 😃

batman

Yes, just recall the dates from database then split them into Year, Moth, Day. Then I'm sure you can just do a simple,

$day1 = "Whatever";
$day2 = "Whatever";
$day 3 = "$day1 - $day2";

Stuff like that.

-Blake

Sphynx

actually, you can just use mktime() to grab the unix timestamp for your start date, and your end date. Then just subtract the starting timestamp from your ending timestamp, and divide by 86400 (second per day). That'll give you the number of days.

epp_b

Yes, do what Sphynx said, it's probably much easier 😃

buck2bcr

This should work for whatever days you put in:

$begin = "2003-02-13";
$end = "2003-03-14";
$start = strtotime($begin);
$finish = strtotime($end);
$div = 60 * 60 * 24;
$days = ($finish - $start)/$div;
echo"Number of days between dates ($begin - $end) = $days";

HTH

Buck