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包含php中2015年有53周的词条

本文目录一览:

php中怎么输出2015年2月份有多少天

用这个函数:int cal_days_in_month ( int $calendar , int $month , int $year )

例子:

?php

$number = cal_days_in_month(CAL_GREGORIAN, 2, 2015); // 28

echo "There were {$number} days in August 2015";

?

PHP计算一年多少个星期和每周的开始和结束日期

PHP计算一年多少个星期和每周的开始和结束日期?方法如下:

方法一:

php

header("Content-type:text/html;charset=utf-8");

date_default_timezone_set("Asia/Shanghai");

$year = (int)$_GET['year'];

$week = (int)$_GET['week'];

$weeks = date("W", mktime(0, 0, 0, 12, 28, $year));

echo $year . '年一共有' . $weeks . '周br /';

if ($week $weeks || $week = 0)

{

$week = 1;

}

if ($week 10)

{

$week = '0' . $week;

}

$timestamp['start'] = strtotime($year . 'W' . $week);

$timestamp['end'] = strtotime('+1 week -1 day', $timestamp['start']);

echo $year . '年第' . $week . '周开始时间戳:' . $timestamp['start'] . 'br /';

echo $year . '年第' . $week . '周结束时间戳:' . $timestamp['end'] . 'br /';

echo $year . '年第' . $week . '周开始日期:' . date("Y-m-d", $timestamp['start']) . 'br /';

echo $year . '年第' . $week . '周结束日期:' . date("Y-m-d", $timestamp['end']);

?

方法二: ?php

header("Content-type:text/html;charset=utf-8");

function getIsoWeeksInYear($year)

{

$date = new DateTime;

$date-setISODate($year, 53);

return ($date-format("W") === "53" ? 53 : 52);

}

function weekday($custom_date)

{

$week_start = date('d-m-Y', strtotime('this week monday', $custom_date));

$week_end = date('d-m-Y', strtotime('this week sunday', $custom_date));

$week_array[0] = $week_start;

$week_array[1] = $week_end;

return $week_array;

}

echo 'br Weeks in 2013br' . getIsoWeeksInYear(2013);

$weekday = weekday(strtotime(date('d-m-Y', strtotime('5-8-2013'))));

echo 'br 10-8-2013';

echo 'brStart: ' . $weekday[0];

echo 'brEnd: ' . $weekday[1];

?

或者方法三:

function get_week($year) {

$year_start = $year . "-01-01";

$year_end = $year . "-12-31";

$startday = strtotime($year_start);

if (intval(date('N', $startday)) != '1') {

$startday = strtotime("next monday", strtotime($year_start)); //获取年第一周的日期

}

$year_mondy = date("Y-m-d", $startday); //获取年第一周的日期

$endday = strtotime($year_end);

if (intval(date('W', $endday)) == '7') {

$endday = strtotime("last sunday", strtotime($year_end));

}

$num = intval(date('W', $endday));

for ($i = 1; $i = $num; $i++) {

$j = $i -1;

$start_date = date("Y-m-d", strtotime("$year_mondy $j week "));

$end_day = date("Y-m-d", strtotime("$start_date +6 day"));

$week_array[$i] = array (

str_replace("-",

".",

$start_date

), str_replace("-", ".", $end_day));

}

return $week_array;

}

函数get_week()通过传入参数$year年份,获取当年第一天和最后一天所在的周数,计算第一周的日期,通过循环获取每一周的第一天和最后一天的日期。最后返回是一个数组。

想得到指定周数的开始日期和结束日期,比如2011年第18周的开始日期和结束日期,代码如下:

复制代码 代码如下:

$weeks = get_week(2011);

echo '第18周开始日期:'.$weeks[18][0].'';

echo '第18周结束日期:'.$weeks[18][1];

最后输出结果:

第18周开始日期:2011.05.02

第18周结束日期:2011.05.08

PHP 根据年、周数获取周的起止日期

我也是来找答案的,结果发现两位仁兄都差了点。

那就让我补上这最后这点吧

function getFirstDayOfWeek($year,$week)

{

$first_day = strtotime($year."-01-01");

$is_monday = date("w", $first_day) == 1;

$week_one_start = !$is_monday ? strtotime("last monday", $first_day) : $first_day;

return $year.'第'.$week.'周开始天:'.date('Y-m-d',$week_one_start+(3600*24*7*($week-1)))

.';结束天为:'.date('Y-m-d',$week_one_start+((3600*24)*(7*($week-1)+6)));

}

PHP 计算某日是这一年的第几周

在判断某一天是哪一年的第几周的时候,根据采用的国际标准(忘了叫什么名字了),年首或者年末的那几天有可能不属于今年的第一周或者最后一周。

代码如下:

?php

echo date("oW",strtotime("20141229"))."\n";

echo date("oW",strtotime('20160101'))."\n";

?

扩展资料

php计算时间段的天数:

$firstday = date("Y-m-d H:i:s",time());//当前日期

$timestamp=strtotime($firstday);//当前日期时间戳

$firstday=date('Y-m-01',strtotime(date('Y',$timestamp).'-'.(date('m',$timestamp)-1).'-01'));//上个月开始的日期

$lastday=date('Y-m-d',strtotime("$firstday +1 month -1 day"));//上个月结束的日期

$stimestamp = strtotime($firstday);

$etimestamp = strtotime($lastday);// 计算日期段内有多少天

$days = ($etimestamp-$stimestamp)/86400+1;// 保存每天日期

$date = array();

for($i=0; $i$days; $i++){

$date[] = date('Y-m-d', $stimestamp+(86400*$i));

}

php 根据多少天算几周 比如15天 2周+1天 这如何算?

$allday = 15;

$weekday = 7;

$hasweek = floor($allday/$weekday);//下取整计算有几个星期

$lessday = $allday%$weekday;//取余计算不足一星期的天数