date operations

Anon

Hello
How can I make date operations?
For example: Add days to a date

Sam

Anon

That's how you can do it:

$today=date('d.m.Y');
$tomorrow = mktime (0,0,0,date("m") ,date("d")+1,date("Y"));
$tomorrow = date('d.m.Y', $tomorrow);

So, today mask means that it will be 'day.month(2 digits).year(4 digits)'. Date formats you can find in PHP manual on this site.

mktime description is:

int mktime(int hour, int minute, int second, int month, int day, int year, int [is_dst] ); -- Get UNIX timestamp for a date.

day() function formats date which you place in it or returns current date if no date is specified.

That's it ;-)

Anon

what about if i want to make date operations on a given date other than today.

i mean i have a variable, let's say $data which can be other than today, and i want to calculate ... let's say $yesterday, with refference to $data.

can mktime do this trick?

thanks...

bbbear

Good day. I have the same problem as you are and I wonder if you've found a solution ? If so, could you please tells me how to do it ?

What I need to do is to add 30 days to a given date. I've been looking for solutions here and there and hopefully you can help me in this.

Thank you in advance.

Anon

Should be possible.

Use:
mktime(Hour, Minute, Second, Month, Day, Year)

For example:
$nTmp = mktime(0,0,0,1,1-1,2001);
echo strftime("%d.%m.%Y",$nTmp);
returns 31.12.2001

Hope that helps
Greetings Sam

Anon

I used the function mktime.

mktime(0, 0, 0, month, day+30, year)
will return a timestamp which contain your date plus 30 days.

Greetings
Sam

lucki

let's say your refference data is

$data="2000-08-15";

so what do you need is to transform a string into a format that php understands as a date.

hopefully strftime() function parses this string into a timestamp (the number of seconds that passed since the unix era -1970-01-01).
then you can actually make whatever operations you want with strftime() which can even use dummy words like "-1 month" (see examples)

it is important when you use strftime to use setlocale() function;

look at these examples

setlocale(LC_TIME, "en_EN");

$data1=strftime("%Y-%m-%d",strtotime("$data -1 month"));

$data3=strftime("%Y-%m-%d",strtotime("$data -3 months"));

$data6=strftime("%Y-%m-%d",strtotime("$data -6 months"));

$data12=strftime("%Y-%m-%d",strtotime("$data -1 year"));

i hope it will help