日期格式不提供有效的日期，即使它匹配文档 - php date formating help Date format does not give valid date even though it matches documentation

php date formating help Date format does not give valid date even though it matches documentation

本文关键字：日期文档格式有效 | 更新日期: 2023-09-27

阅读并重新阅读文档，确实无法解释为什么$date1通过坏

有人能帮忙吗?非常感谢

$date1 = "04-16-2013";
$date2 = "2013-04-16";
printf("date1 = %s'n", $date1);
printf("date2 = %s'n", $date2);
$newdate1 = date('m-d-Y',$date1);
$newdate2 = date('Y-m-d',$date2);
printf("newdate1 = %s'n", $newdate1);
printf("newdate2 = %s'n", $newdate2);
$previous_date = date('m-d-Y', strtotime($date1 .' -1 day'));
$previous_date2 = date('m-d-Y', strtotime($date2 .' -1 day'));
printf("previous_date = %s'n", $previous_date);
printf("previous_date2 = %s'n", $previous_date2);

输出

 date1 = 04-16-2013
    date2 = 2013-04-16
    newdate1 = 12-31-1969
    newdate2 = 1969-12-31
    previous_date = 12-31-1969
    previous_date2 = 04-15-2013

strtotime function

m/d/y或d-m-y格式的日期通过查看来消除歧义各个组件之间的分隔符:如果分隔符是a斜杠(/)，则采用美式的m/d/y;然而，如果分隔符是一个破折号(-)或点(.)，然后是欧洲的d-m-y格式假定.

为了避免潜在的歧义，最好使用ISO 8601 (YYYY-MM-DD)日期或DateTime::createFromFormat()如果可能的话。

在您的情况下，因为您使用分隔符(-)php考虑日期格式为d-m-y。

或者变通方法:

$date1 = "04-16-2013";
$date2 = "2013-04-16";
printf("date1 = %s'n", $date1);
printf("date2 = %s'n", $date2);

$newdate1 = date('m-d-Y',strtotime(str_replace("-","/",$date1)));
$newdate2 = date('Y-m-d', strtotime($date2));
printf("newdate1 = %s'n", $newdate1);
printf("newdate2 = %s'n", $newdate2);

$previous_date = date('m-d-Y', strtotime(str_replace("-","/",$date1) .' -1 day'));
$previous_date2 = date('m-d-Y', strtotime($date2 .' -1 day'));
printf("previous_date = %s'n", $previous_date);
printf("previous_date2 = %s'n", $previous_date2);

在您的情况下试试:

$date1 = "04-16-2013";
$date = date_create_from_format('m-j-Y', $date1);
echo date_format($date, 'd-m-Y');

首先输入月-日-年....然后输出日-月-年