<input type = "date" name = "myDate" onchange = ""> //this is input date
当用户选择一个日期然后它想根据上面的日期进行更改 如何根据给定日期进行更改,请帮助我..!
<table class="table table-bordered" id="thbg">
<thead><td></td>
<th>Project</th>
<th>Activity</th>
<th>Bill Type</th>
<?php
$day = "1";
$month = "11";
$year = "2017";
$start_date = $day."-".$month."-".$year;
$start_time = strtotime($start_date);
$end_time = strtotime("+1 week", $start_time);
for($i=$start_time; $i<$end_time; $i+=86400)
{
print '<th align="center"> '. date("m-d-Y l", $i). '</th>';
}
?>
<th>Hours</th>
</thead>
</table>
并且输出也有错误的日期,例如我一次有两个星期天
谁能提出解决方案?
最佳答案
不确定您在这里尝试实现什么,但更好的选择是使用 DateTime
类
类似的东西应该可以工作
$day = "1";
$month = "11";
$year = "2017";
$objDateStart = DateTimeImmutable::createFromFormat('j-m-Y', $day."-".$month."-".$year);
$objDateEnd = $objDateStart->modify('+1 week');
$objDateRange = new DatePeriod($objDateStart, new DateInterval('P1D'), $objDateEnd);
foreach($objDateRange as $objDate)
{
echo '<th align="center"> '. $objDate->format("m-d-Y l"). '</th>';
}
关于javascript - Codeigniter-PHP : How to Change table week Dates , 更改输入日期,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47571290/