用PHP获取一周的第一天？

Question

鉴于日期MM-dd-yyyy格式,有人可以帮助我获得一周的第一天吗？

Answer 1

这是我正在使用的......

$day = date('w');
$week_start = date('m-d-Y', strtotime('-'.$day.' days'));
$week_end = date('m-d-Y', strtotime('+'.(6-$day).' days'));

$ day包含0到6之间的数字,表示星期几(星期日= 0,星期一= 1等).
$ week_start包含当前星期日的日期为mm-dd-yyyy.
$ week_end包含当前星期六的日期为mm-dd-yyyy.

Answer 2

strtotime('this week', time());

替换时间().当前一天是星期日/星期一,下周日/最后一个星期一方法将无效.

Answer 3

使用方法很简单strtotime函数:

echo date("Y-m-d", strtotime('monday this week')), "\n";   

echo date("Y-m-d", strtotime('sunday this week')), "\n";

它与PHP版本有所不同:

输出为5.3.0 - 5.6.6,php7 @ 20140507 - 20150301,hhvm-3.3.1 - 3.5.1

2015-03-16
2015-03-22

输出为4.3.5 - 5.2.17

2015-03-23
2015-03-22

输出为4.3.0 - 4.3.4

2015-03-30
2015-03-29

Edge-Cases比较

本周的相关描述有自己的背景.下面显示的输出,本周周一,当它是一个星期一或星期天星期天:

$date = '2015-03-16'; // monday
echo date("Y-m-d", strtotime('monday this week', strtotime($date))), "\n";   
echo date("Y-m-d", strtotime('sunday this week', strtotime($date))), "\n";

$date = '2015-03-22'; // sunday
echo date("Y-m-d", strtotime('monday this week', strtotime($date))), "\n";   
echo date("Y-m-d", strtotime('sunday this week', strtotime($date))), "\n";

Againt它与PHP版本有所不同:

输出为5.3.0 - 5.6.6,php7 @ 20140507 - 20150301,hhvm-3.3.1 - 3.5.1

2015-03-16
2015-03-22
2015-03-23
2015-03-29

输出为4.3.5 - 5.0.5,5.2.0 - 5.2.17

2015-03-16
2015-03-22
2015-03-23
2015-03-22

输出为5.1.0 - 5.1.6

2015-03-23
2015-03-22
2015-03-23
2015-03-29

输出为4.3.0 - 4.3.4

2015-03-23
2015-03-29
2015-03-30
2015-03-29

Answer 4

把事情简单化 :

<?php    
$dateTime = new \DateTime('2012-05-14');
$monday = clone $dateTime->modify(('Sunday' == $dateTime->format('l')) ? 'Monday last week' : 'Monday this week');
$sunday = clone $dateTime->modify('Sunday this week');    
?>

资料来源:PHP手册

Answer 5

$givenday = date("w", mktime(0, 0, 0, MM, dd, yyyy));

这将为您提供给定日期本周的某一天,其中0= Sunday和6= Saturday.从那里你可以简单地向后计算你想要的那一天.

Answer 6

对于它的价值,这里是确定一致参考框架时strtotime的不稳定行为的细分:

http://gamereplays.org/reference/strtotime.php

基本上只有这些字符串能够可靠地给你相同的日期,无论你打电话给他们当天的哪一天:

strtotime("next monday");
strtotime("this sunday");
strtotime("last sunday"); 

Answer 7

这个问题需要一个很好的DateTime答案: -

function firstDayOfWeek($date)
{
    $day = DateTime::createFromFormat('m-d-Y', $date);
    $day->setISODate((int)$day->format('o'), (int)$day->format('W'), 1);
    return $day->format('m-d-Y');
}

var_dump(firstDayOfWeek('06-13-2013'));

输出: -

string '06-10-2013' (length=10)

这将涉及年份边界和闰年.

Answer 8

以下代码应适用于任何自定义日期,只使用所需的日期格式.

$custom_date = strtotime( date('d-m-Y', strtotime('31-07-2012')) ); 
$week_start = date('d-m-Y', strtotime('this week last monday', $custom_date));
$week_end = date('d-m-Y', strtotime('this week next sunday', $custom_date));
echo '<br>Start: '. $week_start;
echo '<br>End: '. $week_end;

我用PHP 5.2.17测试了代码结果:

Start: 30-07-2012
End: 05-08-2012

Answer 9

假设星期一是一周的第一天,这有效:

echo date("M-d-y", strtotime('last monday', strtotime('next week', time())));

Answer 10

这个怎么样？

$first_day_of_week = date('m-d-Y', strtotime('Last Monday', time()));
$last_day_of_week = date('m-d-Y', strtotime('Next Sunday', time()));

Answer 11

在这里,我将周日视为本周末的第一天和周六.

$m = strtotime('06-08-2012');  
$today =   date('l', $m);  
$custom_date = strtotime( date('d-m-Y', $m) );   
if ($today == 'Sunday') {  
   $week_start = date("d-m-Y", $m);  
} else {  
  $week_start = date('d-m-Y', strtotime('this week last sunday', $custom_date));  
}  

if ($today == 'Saturday') {  
  $week_end = date("d-m-Y", $m);
} else {  
  $week_end = date('d-m-Y', strtotime('this week next saturday', $custom_date));  
}
echo '<br>Start: '. $week_start;  
echo '<br>End: '. $week_end;  

输出:

开始于:05-08-2012
结束:11-08-2012

Answer 12

这是我用来从任何日期获得一周的第一天和最后一天.在这种情况下,星期一是一周的第一天......

$date = date('Y-m-d') // you can put any date you want
$nbDay = date('N', strtotime($date));
$monday = new DateTime($date);
$sunday = new DateTime($date);
$monday->modify('-'.($nbDay-1).' days');
$sunday->modify('+'.(7-$nbDay).' days');

Answer 13

这个怎么样？

$day_of_week = date('N', strtotime($string_date));
$week_first_day = date('Y-m-d', strtotime($string_date . " - " . ($day_of_week - 1) . " days"));
$week_last_day = date('Y-m-d', strtotime($string_date . " + " . (7 - $day_of_week) . " days"));

Answer 14

只需使用 date($format, strtotime($date,' LAST SUNDAY + 1 DAY'));

Answer 15

尝试这个：

function week_start_date($wk_num, $yr, $first = 1, $format = 'F d, Y')
{
    $wk_ts  = strtotime('+' . $wk_num . ' weeks', strtotime($yr . '0101'));
    $mon_ts = strtotime('-' . date('w', $wk_ts) + $first . ' days', $wk_ts);
    return date($format, $mon_ts);
}

$sStartDate = week_start_date($week_number, $year);
$sEndDate   = date('F d, Y', strtotime('+6 days', strtotime($sStartDate)));

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