我有一个场景，用户选择一个时间和日期（或多天），并且该值必须转换为该日期和时间的 UTC 时间。我有每个用户的 gmt 偏移量（用户在注册时设置它）。例如：
东部时区的用户选择：
周一、周二、周五下午 3:15
我需要知道该信息在 UTC 时间中的时间和日期。解决方案必须考虑到这样的情况，例如一个时区的星期一可能是 UTC 时间的另一天。另外，如果可以将时间转换为 24 小时格式，那将是一个加分项。
为了清楚起见，应该返回类似于数组的内容，例如：
Array('<3:15 pm eastern adjusted for utc>', '<Monday adjusted for UTC>', '<Tuesday adjusted for UTC>', '<Friday adjusted for UTC>');
我不需要将结果直接格式化为这样的数组 - 这只是最终目标。
我猜这涉及到使用 strtotime，但我就是不知道该怎么做。
【问题讨论】：
标签： php datetime time timezone
$timestamp = strtotime($input_time) + 3600*$time_adjustment;
结果将是一个时间戳，这是一个示例：
$input_time = "3:15PM 14th March";
$time_adjustment = +3;
$timestamp = strtotime($input_time) + 3600*$time_adjustment;
echo date("H:i:s l jS F", $timestamp);
// 16:15:00 Monday 14th March
编辑：一直忘记一些小事，现在应该可以正常工作了。
【讨论】：
制作了一个函数来完成这项工作：
<?
/*
* The function week_times() converts a a time and a set of days into an array of week times. Week times are how many seconds into the week
* the given time is. The $offset arguement is the users offset from GMT time, which will serve as the approximation to their
* offset from UTC time
*/
// If server time is not already set for UTC, uncomment the following line
//date_default_timezone_set('UTC');
function week_times($hours, $minutes, $days, $offset)
{
$timeUTC = time(); // Retrieve server time
$hours += $offset; // Add offset to user time to make it UTC time
if($hours > 24) // Time is more than than 24 hours. Increment all days by 1
{
$dayOffset = 1;
$hours -= 24; // Find out what the equivelant time would be for the next day
}
else if($hours < 0) // Time is less than 0 hours. Decrement all days by 1
{
$dayOffset = -1;
$hours += 24; // Find out what the equivelant time would be for the prior day
}
$return = Array(); // Times to return
foreach($days as $k => $v) // Iterate through each day and find out the week time
{
$days[$k] += $dayOffset;
// Ensure that day has a value from 0 - 6 (0 = Sunday, 1 = Monday, .... 6 = Saturday)
if($days[$k] > 6) { $days[$k] = 0; } else if($days[$k] < 0) { $days[$k] = 6; }
$days[$k] *= 1440; // Find out how many minutes into the week this day is
$days[$k] += ($hours*60) + $minutes; // Find out how many minutes into the day this time is
}
return $days;
}
?>
【讨论】：