Date increment

[cvzyl]

Hi

 

I am a newbie to PHP so if this is a really dumb question, I apologise in advance.

 

I have a value stored in variable $date and want to increment this date by one day, something like:

$date = $date + 1;

 

I cannot find a date increment function in the PHP help file.  I thought that this would be quite a commonly used function but maybe I am looking in the wrong place.

 

How should I go about doing this?

 

Thanks in advance

Cobus

[Ninjakreborn]

Turn it into a timestamp (run it through the strtotime function), and then add "UTC days are mostly 86400 s long, but are occasionally 86401 s and could be 86399 s long" - referenced from <a href="http://en.wikipedia.org/wiki/Unix_time">HERE</a> to it

[JBS103]

$date =  strtotime("+1 day");

Will give tommorow based off the current time.

 

$date += 86400;

Will give any $date plus 1 day assuming you are working off seconds.

[Barand]

I have a value stored in variable $date

 

How you add 1 day depends on how that date variable is stored

 

If it's in the form of a date, eg

$date = "2007-03-07" ;

$plus_one_day = date ('Y-m-d', strtotime ("+1 days $date"));

 

If it's a unix timestamp in seconds you can add 86400 as above or

$plus_one_day = date ('Y-m-d', strtotime ("+1 days", $date));

 

The advantage of the strtotime method is the increment can be

 

"+x days"

"+x hours"

"-x days"

 

etc etc withour you having to calculate the number of seconds.

 

[cvzyl]

Thanks for all the comments, I managed to get all the different methods to work but it seems like the suggestion from Barand

$plus_one_day = date ('Y-m-d', strtotime ("+1 days", $date));

is the cleanest solution.

 

My next problem is, how do I "terminate" a date to only contain Y-m-d without any hours or seconds.  I want 2007-03-07 00:00:00 rather than 2007-03-07 23:25:37.  I want to convert a variable (e.g. $date) to this "terminated" value.

 

Cobus

[Barand]

If it's a database field specify as DATE and not DATETIME.

 

BTW, how are you storing that date variable?

[cvzyl]

Barand

 

The database field is defined as DATE and not DATETIME.

 

What do you mean how am I storing the variable?

 

Cobus

[artacus]

Just do it when you are pulling it from the database

SELECT DATE_ADD(myDate, INTERVAL 1 DAY) AS newDay

[Barand]

The database field is defined as DATE and not DATETIME.

 

In that case the time element will be 00:00:00

 

 

What do you mean how am I storing the variable?

 

I meant as a date or unix time