PHP检查两个日期时间是否不在同一日历日 [英] PHP check if two Datetimes are not on the same calendar day

问题描述

我有两个这样的日期时间（日期实际上是$ vars）

I have two Datetimes like this (the dates being actually $vars)

$startTime = \DateTime::createFromFormat('Y/m/d H:i', '2015/01/01 23:00');
$endTime = \DateTime::createFromFormat('Y/m/d H:i', '2015/01/02 01:00');

我遇到了一个（可能很漂亮）简单的问题：如何确定两个日期是否在同一天

I struggle with a (possibly pretty) simple problem: How could I determine if the two dates are on different calendar days?

我不能像 2015/01/01 22那样执行< ：00< 2015/01/01 23:00 也是正确的。我也不能这样做：

I cannot do < as 2015/01/01 22:00 < 2015/01/01 23:00 would also be true. I can also not do this:

$diff = $startTime->diff($endTime);
$days = $diff->format('%d');

echo $days;

，因为它给了我 0 。

THIS 给了我一个有关如何做的想法，但是对于javascript，相当于php的是什么？

THIS gives me an idea about how to do it, but for javascript, what would be the equivalent for php?

// UPDATE

//UPDATE

    $startDate = $startTime->format('Y/m/d');
    $endDate = $endTime->format('Y/m/d');
    $diffDates = $startDate->diff($endDate);
    $daysDiff = $diffDates->format('%d');

    echo $daysDiff;

由于评论，我认为这现在可能是正确的方法，但是现在我得到错误：在字符串上调用成员函数diff（）

I think that might be the right approach now, thanks to the comments, but now I get Error: Call to a member function diff() on string

//为澄清我想做的更新

//UPDATE FOR CLARIFICATION WHAT I'M TRYING TO DO

我只希望天数有所不同，因此对于上面的内容，它应为'1'（尽管实际上仅相差2小时），例如'2015/01/01 23:00'和'2015/01/03 17:00'将是'2'。

I just want to have the difference in days, so for the above it would be '1' (although only 2 hours difference actually) and for example '2015/01/01 23:00' and '2015/01/03 17:00' would be '2'.

推荐答案

只需创建时间设置为00:00:00的日期：

Just create the dates with time set to 00:00:00:

$startTime = \DateTime::createFromFormat('Y/m/d H:i:s', '2015/01/01 00:00:00');
$endTime = \DateTime::createFromFormat('Y/m/d H:i:s', '2015/01/02 00:00:00');

或在现有日期将时间重置为零：

or reset time to zero on existing dates:

$startTime->setTime(0, 0, 0);
$endTime->setTime(0, 0, 0);

然后它应该可以工作：

$diff = $startTime->diff($endTime);
$days = $diff->format('%d');

echo $days; // 1

奖金

如果只想使用日期，请记住在 createFromFormat 中将时间设置为00:00:00或使用进行重置。 setTime 。如果您不会在 createFromFormat 中提供时间，PHP会将其设置为当前时间：

If you want to work only with dates, remember to set the time to 00:00:00 in createFromFormat or reset it with setTime. If you won't provide time in createFromFormat PHP will set it to the current time:

$date = DateTime::createFromFormat('Y-m-d', '2016-01-21');
print $date->format('H:i:s'); //not 00:00:00

要解决此问题，您必须：

To fix it, you must either:

• 以格式提供00:00:00时间：

• provide 00:00:00 time in format:

$ date = DateTime :: createFromFormat（'Ymd H：i：s'，'2016-01-21 00:00:00'）;

在日期格式前加上感叹号并省略时间，这将自动将时间设置为00:00:00：

prefix the date format with exclamation mark and omit the time, this will set the time to 00:00:00 automatically:

$ date = DateTime :: createFromFormat（'！Ym-d'，'2016-01-21'）;

重置创建后的时间：

$ date = DateTime :: createFromFormat（'Ym-d'，'2016-01-21' ）;
$ date-> setTime（0，0）;

这篇关于PHP检查两个日期时间是否不在同一日历日的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！