count how many day fall into specific date a range of year

Question

Initially I am trying to create a function to display how many times does a specific day fall into particular date. for example how many times does Saturday fall into January 1st of certain year to certain year.

<?php 

$firstDate = '01/01/2000';
$endDate = '01/01/2012';
$newYearDate= '01/01';

# convert above string to time
$time1 = strtotime($firstDate);
$time2 = strtotime($endDate);
$newYearTime = strtotime($newYearDate);

for($i=$time1; $i<=$time2; $i++){
    $saturday = 0;
    $chk = date('D', $newYearTime); #date conversion
   if($chk == 'Sat' && $chk == $newYearTime){ 
      $saturday++;    
      } 
}
echo $saturday;

?>

Answer 1

You can only have a saturday to be in, say January 1, once in a year, so:

$firstDate = '01/01/2000';
$endDate = '01/01/2012';

$time1 = strtotime($firstDate);
$time2 = strtotime($endDate);

$saturday = 0;
while ($time1 < $time2) {

    $time1 = strtotime(date("Y-m-d", $time1) . " +1 year");
    $chk = date('D', $time1);
    if ($chk == 'Sat') {
        $saturday++;
    }

}

echo "Saturdays at 01/01/yyyy: " . $saturday . "\n";

The line I changed was:

$time1 = strtotime(date("Y-m-d", strtotime($time1)) . " +1 year");

to

$time1 = strtotime(date("Y-m-d", $time1) . " +1 year");

as $time1 is already in seconds from the epoch -- the format required for date.

Answer 2

strtotime gives you seconds since 1970-01-01. Since you're interested in days only, you can increment your loop by 86400 seconds per day to speed up your calculation

for($i = $time1; $i <= $time2; $i += 86400) {
...
}

There are several points

• move $saturday out of your loop
• check new year's eve with day of year
• check the loop counter $i instead of $newYearTime

This should work

$firstDate = '01/01/2000';
$endDate = '01/01/2012';

# convert above string to time
$time1 = strtotime($firstDate);
$time2 = strtotime($endDate);

$saturday = 0;
for($i=$time1; $i<=$time2; $i += 86400){
    $weekday = date('D', $i);
    $dayofyear = date('z', $i);
    if($weekday == 'Sat' && $dayofyear == 0){ 
        $saturday++;    
    } 
}

echo "$saturday\n";