[SOLVED] How can I use PHP to list all the dates in a year?

How can I use PHP to list all the dates in a year?

I'm trying to use PHP to create a script that searches all the days between now and one year's time and lists all the dates for Fridays and Saturdays. I was trying to use PHP's date() and mktime() functions but can't think of a way of doing this. Is it possible?

Thanks, Ben

Solution

Here's how to do it in a cool way, with special thanks to strtotime's relative formats.

$friday = strtotime('Next Friday', time());
$saturday = strtotime('Next Saturday', time());
$friday = strtotime('+1 Week', $friday);
$saturday = strtotime('+1 Week', $saturday);

Of course you should tweak it to do exactly what you want, but that's beside the point I was trying to make.

Also note that strtotime will give you timestamps. To find out the date use:

date('Y-m-d', $friday)

Another thing to know is that Next <dayofweek> will exclude your current day from the search, so if you also want to include the current day you can do it like this:

$friday = strtotime('Next Friday', strtotime('-1 Day', time()));

And here's a full working script that does exactly what you wanted.

<?php
// prevent multiple calls by retrieving time once //
$now = time();
$aYearLater = strtotime('+1 Year', $now);

// fill this with dates //
$allDates = Array();

// init with next friday and saturday //
$friday = strtotime('Next Friday', strtotime('-1 Day', $now));
$saturday = strtotime('Next Saturday', strtotime('-1 Day', $now));

// keep adding days untill a year has passed //
while(1){
    if($friday > $aYearLater)
        break 1;
    $allDates[] = date('Y-m-d', $friday);
    if($saturday > $aYearLater)
        break 1;
    $allDates[] = date('Y-m-d', $saturday);

    $friday = strtotime('+1 Week', $friday);
    $saturday = strtotime('+1 Week', $saturday);
}

//XXX: debug
var_dump($allDates);

?>

Good luck, Alin