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@NogDogSep 16.2009 — #$tomorrow would be an integer value defining a number of seconds since the "UNIX Epoch" (1970/01/01-00:00:00GMT), a.k.a. a UNIX timestamp. When date() is given only one argument -- the format string -- it uses the current system time as that timestamp. In your example, the preceding line is going through a bunch of contortions to define $tomorrow to be the UNIX Timestamp integer for "tomorrow", then that value is used as the 2nd argument to date(). It would be a bit simpler to use strtotime() to get "tomorrow":
@NogDogSep 16.2009 — #
Yes
Yes, assuming that $tomorrow has been assigned a timestamp value for tomorrow's date, as your sample code does via mktime().
[url=http://php.net/strtotime]strtotime[/url] () is much easier for something like that:
However, this by itself is silly code by itself, since all you would need to do to output a specific date is to just output it:
For learning purposes, though, it could be supposed that the timestamp value might have been derived in some programmatic manner and not simply assigned a hard-coded value.
learning PHP
[url]http://www.w3schools.com/PHP/php_date.asp
[CODE]PHP Date() – Adding a Timestamp
The optional timestamp parameter in the date() function specifies a timestamp. If you do not specify a timestamp, the current date and time will be used.
The mktime() function returns the Unix timestamp for a date.
The Unix timestamp contains the number of seconds between the Unix Epoch (January 1 1970 00:00:00 GMT) and the time specified.
Syntax for mktime()
mktime(hour,minute,second,month,day,year,is_dst)
To go one day in the future we simply add one to the day argument of mktime():
<?php
$tomorrow = mktime(0,0,0,date(“m”),date(“d”)+1,date(“Y”));
echo “Tomorrow is “.date(“Y/m/d”, $tomorrow);
?>
The output of the code above could be something like this:
Tomorrow is 2009/05/12
I dont understand the output .
let me analyse it
we have print of date(“Y/m/d”, $tomorrow);
“Y/m/d” –>gives 2009/05/12.
Now there is another part i.e $tomorrow …..[B][COLOR=”Blue”]where is the print for $tomorrow? and if its not for printing then whats the role of of this $tomorrow here ? i am exactly stuck here .
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@NogDogSep 16.2009 — #$tomorrow would be an integer value defining a number of seconds since the "UNIX Epoch" (1970/01/01-00:00:00GMT), a.k.a. a UNIX timestamp. When date() is given only one argument -- the format string -- it uses the current system time as that timestamp. In your example, the preceding line is going through a bunch of contortions to define $tomorrow to be the UNIX Timestamp integer for "tomorrow", then that value is used as the 2nd argument to date(). It would be a bit simpler to use strtotime() to get "tomorrow":[code=php]
$tomorrow = strtotime('tomorrow');
echo date('Y/m/d', $tomorrow);
// or even:
echo date('Y/m/d', strtotime('tomorrow'));
[/code]0
@NogDogSep 16.2009 — #Hi,I appreciate for the alternative suggestion/examples.
by the way, i become totally confused because you told lot many things which are bouncers.
in simple way, if i understand you correctly ...
(1) date('Y/m/d') prints the current system date , time of my machine. ....correct ?
[/quote]
Yes
(2)if i put date('Y/m/d', $tomorrow) does this means ...its a tomorrow date ?
[/quote]
Yes, assuming that $tomorrow has been assigned a timestamp value for tomorrow's date, as your sample code does via mktime().
How about if we want to make a date say 29th September 2009 ...do we still be using mktime() function ?
If thats so, then how could i make a absolute date say 29th September 2009 using mktime() ....and print it ?[/QUOTE]
[code=php]
$some_day = strtotime('2009-09-29'); // unix timestamp integer
echo date('Y/m/d', $some_day);
[/code]However, this by itself is silly code by itself, since all you would need to do to output a specific date is to just output it:
[code=php]
echo "2009/09/29";
[/code]For learning purposes, though, it could be supposed that the timestamp value might have been derived in some programmatic manner and not simply assigned a hard-coded value.