问题描述

我正在从数据库中检索格式为 2013-09-15 08:45:00 的日期，该日期设置为 UTC，我需要将其更改为另一个动态时区(基于用户)

I am retrieving a date in format of 2013-09-15 08:45:00 from the database, which is set in UTC and I need to change it to another dynamic timezone (based on user)

到目前为止我有

$datetime = $row->field_data_field_performance_times_field_performance_times_v;
$eventDate = DateTime::createFromFormat('Y-m-d H:i:s', $datetime, new DateTimeZone($user->timezone));
$performance_time = date_format($eventDate, 'l, j F, Y, H:i');

但它不会改变时区.任何想法有什么问题?就我而言，应该是 +2 小时.

But it doesn't change the timezone. Any ideas what's wrong? It should be +2 hours in my case.

推荐答案

您输入的日期时间是 UTC，而不是用户的时区.因此，首先您必须在 UTC 中创建日期时间对象，然后将时区设置/更改为用户的 :

Your input datetime is in UTC, not user's timezone. So first you must create datetime object in UTC, and then set/change timezone to user's :

$dt = new DateTime('2013-09-15 08:45:00', new DateTimeZone('UTC'));
print_r($dt);
/*
DateTime Object
(
    [date] => 2013-09-15 08:45:00
    [timezone_type] => 3
    [timezone] => UTC
)
*/

现在您有 UTC 时区的日期时间.如果您想更改时区，只需在 DateTime 对象上调用 ->setTimezone() :

Now you have datetime in UTC timezone. If you wish to change timezone, just call ->setTimezone() on DateTime object :

$dt->setTimezone(new DateTimeZone('Europe/Berlin'));
print_r($dt);
/*
DateTime Object
(
    [date] => 2013-09-15 10:45:00
    [timezone_type] => 3
    [timezone] => Europe/Berlin
)
*/

附言因为输入 2013-09-15 08:45:00 是标准的日期时间格式，所以你不需要使用 DateTime::createFromFormat.

p.s. because input 2013-09-15 08:45:00 is in standard datetime format, you don't need to use DateTime::createFromFormat.

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