使用php以10次迭代一周递增一个日期
我想将开始日期增加1周10次。此代码不起作用:使用php以10次迭代一周递增一个日期,php,Php,我想将开始日期增加1周10次。此代码不起作用: <?php $start_date = "06/25/2012"; $date = strtotime($start_date); $X=1; while ($X <= 10) {$X++; $Y=7*$X; $date = strtotime("+ $Y days", $date); echo date('m/d/Y', $date)."<br>"; } 这是错误的 您可能正在寻找以下内容: $start\
<?php
$start_date = "06/25/2012";
$date = strtotime($start_date);
$X=1;
while ($X <= 10) {$X++; $Y=7*$X;
$date = strtotime("+ $Y days", $date);
echo date('m/d/Y', $date)."<br>";
}
这是错误的 您可能正在寻找以下内容:
$start\u date=“2012年6月25日”;
$date=strottime($start\u date);
$X=1;
而($X使用DateTime
对象:
$date = new DateTime('2012-06-25');
for ($i = 1; $i <= 10; $i++) {
$date->modify('+1 week');
echo $date->format('m/d/Y');
}
$date=新日期时间('2012-06-25');
对于($i=1;$i修改(“+1周”);
echo$date->format('m/d/Y');
}
Ideone:试试这段代码。这将为您提供50次1周周期日期列表
<?php
$date = "05/01/2016";
$weekDates = '';
for ( $i = 1; $i <= 50; $i++ ) {
$date = date("m/d/Y", strtotime("+1 weeks", strtotime($date)));
$weekDates .= $date . '<br>';
}
echo $weekDates;
回答:
<?php
$start_date = "06/25/2012";
$date = strtotime($start_date);
$newDate = $date;
for ($x = 1; $x<=10; $x++)
{
$newDate = $newDate + (86400*7);//adding the amount of seconds in a week
$buffer = date("n/j/Y", $newDate);
echo "{$buffer}<br>";
}
试试这个回音日期(“m.d.Y”,strotime(+1周),strotime($start_date))
把$Y
改成7
$date=strotime(+7天),日期));告诉我我的答案是否对你有用。其他答案也都是正确的。但这个答案遵循了我提供的代码。
<?php
$date = "05/01/2016";
$weekDates = '';
for ( $i = 1; $i <= 50; $i++ ) {
$date = date("m/d/Y", strtotime("+1 weeks", strtotime($date)));
$weekDates .= $date . '<br>';
}
echo $weekDates;
<?php
$start_date = "06/25/2012";
$date = strtotime($start_date);
$newDate = $date;
for ($x = 1; $x<=10; $x++)
{
$newDate = $newDate + (86400*7);//adding the amount of seconds in a week
$buffer = date("n/j/Y", $newDate);
echo "{$buffer}<br>";
}