How Do You Find The NEXT Year A Day Falls On A Date?

[jjgilels]

Start with: Saturday August 13 1978.  What is the next year that August 13th falls on a Saturday?

 

Couldn't find an answer to that anywhere!

[jcbones]

August 13, 1978 was a Sunday.  This will get the info, that the next time August 13 happens on Sunday is in 1989.  There may be a better way, but this is what I came up with.

 

<?php
$date = new DateTime('1978-08-13');
$findDate = $date->format('l');
$interval = $date->add(new DateInterval('P1Y'));
while($date->format('l') != $findDate) {
$date->add(new DateInterval('P1Y'));
}
echo $date->format('l F d, Y');
?>

[jjgilels]

Where did you run that code?  On my machine, I get this error:

 

Fatal error: Call to undefined method DateTime::add()

 

Jerry.

[jcbones]

http://www.php.net/manual/en/datetime.installation.php

Note: Experimental DateTime support in PHP 5.1.x

Although the DateTime class (and related functions) are enabled by default since PHP 5.2.0, it is possible to add experimental support into PHP 5.1.x by using the following flag before configure/compile: CFLAGS=-DEXPERIMENTAL_DATE_SUPPORT=1

[jcbones]

I did it both ways. So you could use either one.

<?php
//Object Oriented.
echo 'Object Oriented.<br />';
$date = new DateTime('1978-08-13');
$findDate = $date->format('l');
$interval = $date->add(new DateInterval('P1Y'));
while($date->format('l') != $findDate) {
$date->add(new DateInterval('P1Y'));
}
echo $date->format('l F d, Y');

//Procedural
echo '<br /><br />Procedural<br/>';
$date = strtotime('1978-08-13');
$findDate = date('l',$date);
$interval = strtotime('+1 Year',$date);
$current = date('l',$interval);
while($current != $findDate) {
$interval = strtotime('+1 Year',$interval);
$current = date('l',$interval);
}
echo date('l F d, Y',$interval);
?>

[jjgilels]

OK.  The second one works.

 

So, how can I get the NEXT result after today for that 1978 date?

 

Logically, the next time Aug 13 appears on a Sunday after today.  This is actually what I'm after; not the one in 1989.

 

 

[jcbones]

Try this:

$month = '08';
$day = '13';
$year = '1978';
$date = strtotime($year.'-'.$month.'-'.$day);
$findDate = date('l',$date);
$interval = strtotime('+1 Year',strtotime(date('Y').'-'.$month.'-'.$day));
$current = date('l',$interval);
while($current != $findDate) {
$interval = strtotime('+1 Year',$interval);
$current = date('l',$interval);
}
echo date('l F d, Y',$interval);

[jjgilels]

Excellent.  Perfect.  Brilliant.  Suh-weeeet!

 

Thanks.