Question

我正在开发一个php应用程序，它要求我在两个日期之间提取周末的日期，然后将它们作为单个记录插入到mysql数据库中。

我在想是否有更简单的方法，而不是经历开始日期和结束日期之间的循环，以及每个日期检查date('l', strtotime($date))是否返回“星期六”或“星期日”

感谢您的时间

苏尼

Answer 1

如果您使用较旧的PHP安装（或没有DateTime类）：

$now = strtotime("now");
$end_date = strtotime("+3 weeks");

while (date("Y-m-d", $now) != date("Y-m-d", $end_date)) {
    $day_index = date("w", $now);
    if ($day_index == 0 || $day_index == 6) {
        // Print or store the weekends here
    }
    $now = strtotime(date("Y-m-d", $now) . "+1 day");
}

我们遍历日期范围并检查该日期是0还是6索引（星期日或星期六）。

Answer 2

那么php日期（“N”）会给你一周中的一天，其中1表示星期一，7表示星期日，所以你可以很容易地将它变成if语句。

Answer 3

$now = new DateTime("now");
$now->setTime(0,0);
if (($now->format("l") == "Saturday") || ($now->format("l") == "Sunday"))
    $d = $now;
else
    $d = new DateTime("next saturday");

$oneday = new DateInterval("P1D");
$sixdays = new DateInterval("P6D");
$res = array();
while ($d->getTimestamp() <= $endTimestamp) {
    $res[] = $d->format("Y-m-d");
    $d = $d->add($oneday);
    if ($d->getTimestamp() <= $endTimestamp) {
        $res[] = $d->format("Y-m-d");
    }
    $d = $d->add($sixdays);
}

示例

$end = new DateTime("2010-08-20");
$endTimestamp = $end->getTimestamp();

array(6) {
  [0]=>
  string(10) "2010-07-31"
  [1]=>
  string(10) "2010-08-01"
  [2]=>
  string(10) "2010-08-07"
  [3]=>
  string(10) "2010-08-08"
  [4]=>
  string(10) "2010-08-14"
  [5]=>
  string(10) "2010-08-15"
}

Answer 4

这没有经过充分测试，但似乎工作得很好。请注意，如果$ start是在一个周末而$ end是在一周的同一天但是更早，则$ end表示的日期将被省略 - 因此时间戳应该理想地在午夜。

<?php
$start = $now = 1280293200; // starting time
$end = 1283014799; // ending time
$day = intval(date("N", $now));
$weekends = array();
if ($day < 6) {
  $now += (6 - $day) * 86400;
}
while ($now <= $end) {
  $day = intval(date("N", $now));
  if ($day == 6) {
    $weekends[] += $now;
    $now += 86400;
  }
  elseif ($day == 7) {
    $weekends[] += $now;
    $now += 518400;
  }
}
echo "Weekends from " . date("r", $start) . " to " . date("r", $end) . ":\n";
foreach ($weekends as $timestamp) {
  echo date("r", $timestamp) . "\n";
}

Answer 5

For anyone else that finds this - I have used the following:

$start = new DateTime($this->year.'-'.$month.'-01');
$interval = new DateInterval('P1D');
$end = new DateTime($this->year.'-'.$month.'-31');

$period = new DatePeriod($start,$interval,$end);

foreach ($period as $day){
    if ($day->format('N') == 6 || $day->format('N') == 7){
        $compulsory[$day->format('d')] = true;
    }
}

Amend to your own start range and end range and rather than format just the d in $compulsory - you could do d/m/Y for a more substantial date.

在php中获取周末日期

5 个答案: