计算有多少天落入特定日期范围内的年份

Question

最初我试图创建一个函数来显示某一特定日期落入特定日期的次数。例如，星期六在某年某年的1月1日落幕多少次。

<?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

strtotime给你秒自1970-01-01。既然你感兴趣的只有几天，你可以每天86400秒增加你的循环，加快你的计算

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

有几点

• 移动$saturday你的循环
• 检查元旦前夕
• 检查循环计数器$i而不是$newYearTime

这应该工作

$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"; 

Answer 2

只能有一个星期六在，说1月1日，曾经在一年内，所以：

$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"; 

我改了行是：

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

到

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

因为$time1已经从时代开始秒 - 日期所需的格式。