Pull a date and add a few days

[saltious]

Hi Guys.

I need to be able to pull a date and add a few days to it and put it back.

i can already add a few days if the start date is today. however, i have been unable to figure out how to do this if the date that needs adding to is already in the future.

anyone have any ideas?

many thanks.

[Installer]

Use strtotime().

[scrupul0us]

$currentDate = strtotime($pulled_date);

$furtherDate = strtotime(+1 month) + $currentDate;

strtotime() returns a timestamp

[Installer]

strtotime() returns a timestamp

...as is detailed in the linked-to manual page.

Usage:

$currentDate = strtotime($pulled_date); 
$furtherDate = strtotime('+1 month', $currentDate);

[scrupul0us]

ohhh even sleeker usage... i am pwned 😉

[Installer]

Slightly sleeker:

$furtherDate = strtotime($pulled_date . ' +1 month');

[scrupul0us]

careful... you might pwn yourself... theres no telling what happens then

[Installer]

Just trying to give op some code that will work, but thanks for the warning.

[saltious]

thanks for this. i think im getting there. just one more question.

what format does the variable has to be that goes in the brackets?

ie..... $furtherDate = strtotime($pulled_date . ' +1 month');

the $pulled_date variable?

many thanks

[Installer]

I've been assuming you were pulling from a mysql database. If so then data from a DATETIME, DATE, or TIMESTAMP column should work. If not follow the Date Input Formats link from the strtotime manual page.

[saltious]

When i run the function the date im puting into it is in the format 2008/09/13 (for example). then when it goes back into the database with this code.....

$furtherDate = strtotime($daydue . ' +$count2 days');
$completeby2 = date("Y/m/d", $furtherDate);

it goes in as 1970-01-01

what going on, anyone know?

cheers!

[JVassie]

saltious;10885986 wrote:

When i run the function the date im puting into it is in the format 2008/09/13 (for example). then when it goes back into the database with this code.....

$furtherDate = strtotime($daydue . ' +$count2 days');
$completeby2 = date("Y/m/d", $furtherDate);

it goes in as 1970-01-01

what going on, anyone know?

cheers!

That is because (I believe) your input date, 2008/09/13 is not a valid input format for the strtotime() function. Ideally, needs to be in the standard MySQL format which is YYYY-MM-DD.

Try inputting a date in that format and see if it works 🙂

[Installer]

"2008/09/13" is a valid input date (although it's not a MySQL date type, if that's what you're using). Your php code is wrong. I'm not sure what you're trying to do (what is $count?), but this works:

$daydue = '2008/09/13';
$furtherDate = strtotime($daydue . ' +2 days'); 
$completeby2 = date("Y/m/d", $furtherDate);
echo $completeby2;
// echos "2008/09/15"

[saltious]

Hi Thanks, $count2 is simply a number either 0,1 or 2.

i need this to add onto the days.

this is odd. when i output $daydue via the following ($duedate) i get this 2008/09/11. so i know this bit is working

$daydue = mktime(0,0,0,date("m"),date("d"),date("Y"));
$duedate = date("Y/m/d", $daydue);

daydue by itself gives me this. 1221346800. i figure this correct because of the output of the date its gives via duedate. (the numbers by themselves mean nothing to me)

however when i then take this and do the following.

$furtherDate = strtotime($daydue . ' +$count2 days');
$completeby2 = date("Y/m/d", $furtherDate);

it get 1970-01-01. even if i swap the $count2 for just 1 or 2 etc.

i am using mysql by the way. does this change things?

many thanks.

[Installer]

This code is incorrect:

$furtherDate = strtotime($daydue . ' +$count2 days'); 
$completeby2 = date("Y/m/d", $furtherDate);

'+$count2 days' just will not work. Do this:

$furtherDate = strtotime($daydue . ' +' . $count2 . ' days');

If you're storing the date in MySQL as a DATE, DATETIME, or TIMESTAMP type, use hyphens instead of slashes ("2008-09-13").

Edit: You can also do it this way:

$furtherDate = strtotime("$daydue +$count2 days");

although I prefer the other way. They're equally valid and produce the same result. Notice the quote mark differences; remember that variables aren't expanded inside of single quotes but are expanded inside double quotes.

[saltious]

thanks again.

but the bloody thing still isnt working.

any more ideas?

cheers!

[Installer]

daydue by itself gives me this. 1221346800. i figure this correct because of the output of the date its gives via duedate. (the numbers by themselves mean nothing to me)

"1221346800" is what's call a "Unix timestamp". It's merely the number of seconds since "the epoch", i.e. 1970-01-01 00:00:00 UTC, not including leap seconds. You can see how it works with this:

echo strtotime('2008-09-15 09:09:00') . '<br />';
// 1221487740
echo strtotime('2008-09-15 09:09:01') . '<br />';
// 1221487741

[Installer]

but the bloody thing still isnt working.

any more ideas?

Then you're still doing something wrong. I'd have to see your current code.

Paste this code into your editor, save it, run it. It will work:

$count2 = 2;
$daydue = '2008/09/13';
$furtherDate = strtotime("$daydue +$count2 days"); 
$completeby2 = date("Y/m/d", $furtherDate);
echo $completeby2;
// echos "2008/09/15"

[saltious]

this was from user installer

In this line:
PHP Code:
$furtherDate = strtotime($daydue . ' +' . $count2 . ' days');

"$daydue" should be "$duedate" and "$count2" should be "$adddays".

[saltious]

right i am completely lost by this now but ill try.

the count2 is a second adding of days. ill explain this now.

and i added my last response to the wrong sodding thread so things are a little messed up now. ill start again by posting this. it might help

heres my full code for it.

$adddays = $count + $dueby; //count and dueby are 2 seperate numbers between 0-2. from what i can tell this bit works. they each have there own reason and are both needed.

then what i do is run an if statement and if the daydue is a saturday add two days, if its a sunday add one. this is set as $count2

$daydue = mktime(0,0,0,date("m"),date("d")+$adddays,date("Y"));
$duedate = date("Y/m/d", $daydue);

$furtherDate = strtotime($daydue . ' +' . $count2 . ' days');
$completeby2 = date("Y/m/d", $furtherDate);

then for some reason this $completeby2 = 1970-01-01.

anyone any ideas?

thanks for your help by the way, much appreciated