Php 两个时间变量,确定第二个时间是否在第二天
我只有一次约会,两次(24小时制) 有没有一种优雅的方式来判断第二次是否进入第二天 伪代码:Php 两个时间变量,确定第二个时间是否在第二天,php,datetime,Php,Datetime,我只有一次约会,两次(24小时制) 有没有一种优雅的方式来判断第二次是否进入第二天 伪代码: $inputDate1 = "2020-12-27" $inputHour1 = "21:00"; $inputHour2 = "00:10"; $calulatedDate2 = ??? $dateTime1 = $inputDate . " " . $inputHour1; $dateTime2 = $calulat
$inputDate1 = "2020-12-27"
$inputHour1 = "21:00";
$inputHour2 = "00:10";
$calulatedDate2 = ???
$dateTime1 = $inputDate . " " . $inputHour1;
$dateTime2 = $calulatedDate2 . " " . $inputHour1;
其他例子:
08:00和12:00是同一天
08:00和07:59将是下一个日期。检查时间是否小于输入1
$inputDate1 = "2020-12-27";
$inputHour1 = "21:00";
$inputHour2 = "00:10";
if(strtotime($inputHour2) < strtotime($inputHour1)){
$calulatedDate2 = date('Y-m-d', strtotime($inputDate1)+86400);
}else{
$calulatedDate2 = $inputDate1;
}
$dateTime1 = $inputDate1 . " " . $inputHour1;
$dateTime2 = $calulatedDate2 . " " . $inputHour2;
Var_dump($dateTime1, $dateTime2);
$inputDate1=“2020-12-27”;
$inputHour1=“21:00”;
$inputHour2=“00:10”;
if(strotime($inputHour2)
检查时间是否小于输入1
$inputDate1 = "2020-12-27";
$inputHour1 = "21:00";
$inputHour2 = "00:10";
if(strtotime($inputHour2) < strtotime($inputHour1)){
$calulatedDate2 = date('Y-m-d', strtotime($inputDate1)+86400);
}else{
$calulatedDate2 = $inputDate1;
}
$dateTime1 = $inputDate1 . " " . $inputHour1;
$dateTime2 = $calulatedDate2 . " " . $inputHour2;
Var_dump($dateTime1, $dateTime2);
$inputDate1=“2020-12-27”;
$inputHour1=“21:00”;
$inputHour2=“00:10”;
if(strotime($inputHour2)
我找到了一种方法。请随意发布更优雅的解决方案
$inputDate1 = "2020-12-27"
$inputHour1 = "21:00";
$inputHour2 = "00:10";
$inputDate2 = $inputDate1 //just assume for now
$dateTime1 = $inputDate1 . " " . $inputHour1;
$dateTime2 = $inputDate2 . " " . $inputHour1;
if (strtotime($dateTime2) < strtotime($dateTime1)) {
$dateTime2 = (DateTime::createFromFormat('Y-m-d H:i', $dateTime2))->modify('+1 day')->format('Y-m-d H:i');
}
$inputDate1=“2020-12-27”
$inputHour1=“21:00”;
$inputHour2=“00:10”;
$inputDate2=$inputDate1//现在就假设一下
$dateTime1=$inputDate1。" " . $inputHour1;
$dateTime2=$inputDate2。" " . $inputHour1;
if(strotime($dateTime2)修改('+1天')->格式('Y-m-DH:i');
}
我找到了一种方法。请随意发布更优雅的解决方案
$inputDate1 = "2020-12-27"
$inputHour1 = "21:00";
$inputHour2 = "00:10";
$inputDate2 = $inputDate1 //just assume for now
$dateTime1 = $inputDate1 . " " . $inputHour1;
$dateTime2 = $inputDate2 . " " . $inputHour1;
if (strtotime($dateTime2) < strtotime($dateTime1)) {
$dateTime2 = (DateTime::createFromFormat('Y-m-d H:i', $dateTime2))->modify('+1 day')->format('Y-m-d H:i');
}
$inputDate1=“2020-12-27”
$inputHour1=“21:00”;
$inputHour2=“00:10”;
$inputDate2=$inputDate1//现在就假设一下
$dateTime1=$inputDate1。" " . $inputHour1;
$dateTime2=$inputDate2。" " . $inputHour1;
if(strotime($dateTime2)修改('+1天')->格式('Y-m-DH:i');
}
您能检查一下第二次是在第一次之前还是之后吗?这很简单。你能检查一下第二次是在第一次之前还是之后吗?这很简单。哈哈,你赢了我4分钟。哈哈,你赢了我4分钟。