example : my birthday is "may 06 1988" and now "october 02 2007". How many of my age days?
help me plzzz finish my homework.
Full Version: Calculate Date
How to Calculate date with php script?
example : my birthday is "may 06 1988" and now "october 02 2007". How many of my age days?
help me plzzz finish my homework.
example : my birthday is "may 06 1988" and now "october 02 2007". How many of my age days?
help me plzzz finish my homework.
http://php.net/mktime
http://php.net/time
Use those to calculate your age in seconds, then it's just calculation from there down (hint: find years/months/days in seconds and then do some iterative subtraction). I would provide the function I use for this sort of thing, but I'm not going to give you the answer verbatim when it's for homework.
http://php.net/time
Use those to calculate your age in seconds, then it's just calculation from there down (hint: find years/months/days in seconds and then do some iterative subtraction). I would provide the function I use for this sort of thing, but I'm not going to give you the answer verbatim when it's for homework.
Specific questions? Where did I lose you? Lots of smilies, as I have already said to you, don't make a post.
I use this script to calculate days, i m not sure about this
Anyone have more simple than this or correct this?
help me.
CODE
<?php
function CalDate($date1, $date2) {
$stardate = getdate(strtotime($date1));
$enddate = getdate(strtotime($date2));
$kMonth = $stardate[mon];
$kDay = $stardate[mday];
$kYear = $stardate[year];
$d = 0;
$j = $enddate[year] - $stardate[year];
if ($j > 0) {
$c = getdate(strtotime($kYear ."-12-31"));
$d = $c[yday] - $stardate[yday];
for ($i=1;$i<$j;$i++) {
$c = getdate(strtotime($kYear+1 ."-12-31"));
$d = $d + $c[yday] + 1;
}
return $d + $enddate[yday];
}
else return $stardate[yday]-$enddate[yday];
}
print CalDate("2008-01-01","2010-01-02");
?>
function CalDate($date1, $date2) {
$stardate = getdate(strtotime($date1));
$enddate = getdate(strtotime($date2));
$kMonth = $stardate[mon];
$kDay = $stardate[mday];
$kYear = $stardate[year];
$d = 0;
$j = $enddate[year] - $stardate[year];
if ($j > 0) {
$c = getdate(strtotime($kYear ."-12-31"));
$d = $c[yday] - $stardate[yday];
for ($i=1;$i<$j;$i++) {
$c = getdate(strtotime($kYear+1 ."-12-31"));
$d = $d + $c[yday] + 1;
}
return $d + $enddate[yday];
}
else return $stardate[yday]-$enddate[yday];
}
print CalDate("2008-01-01","2010-01-02");
?>
Anyone have more simple than this or correct this?
help me.
If you didn't write that and the you expect to hand it in as 'homework', you should be ashamed of yourself. 
Alex already gave you the closest to an answer, have you actually read his response?
In layman terms:
- Get current UNIX timestamp with http://www.php.net/time
- Get your birthday timestamp with http://www.php.net/mktime
- Do a subtraction of current - yourbirthday, this is the difference in seconds
- Divide by 86400 (60 * 60 * 24 -- seconds in a day) and floor this value http://www.php.net/floor
That's the difference in days.
http://en.wikipedia.org/wiki/Unix_time
If you did, I apologise.
If you did, then I think you should adopt arithmetic over string matching.
Alex already gave you the closest to an answer, have you actually read his response?
In layman terms:
- Get current UNIX timestamp with http://www.php.net/time
- Get your birthday timestamp with http://www.php.net/mktime
- Do a subtraction of current - yourbirthday, this is the difference in seconds
- Divide by 86400 (60 * 60 * 24 -- seconds in a day) and floor this value http://www.php.net/floor
That's the difference in days.
http://en.wikipedia.org/wiki/Unix_time
If you did, I apologise.
If you did, then I think you should adopt arithmetic over string matching.
all this stuf make me confuse but thanks guys for help me
i will learn more
i will learn more
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