Question

我有一个场景，用户选择时间和日期（或多天），并且该值必须转换为UTC时间中那天和时间。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

我需要知道在UTC时间内信息的时间和天数。解决方案必须考虑星期一的情况，在一个时区可以是UTC时间的不同日期。此外，如果时间可以转换为24小时格式，那将是一个加号。

为了清楚起见，应该返回数组中的某些内容，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。

我猜它涉及使用strtotime，但我不能完全指责如何去做。

Answer 1

 $timestamp = strtotime($input_time) + 3600*$time_adjustment;

结果将是一个时间戳，这是一个例子：

 $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：一直忘记一些小事情，现在应该做得很好。

Answer 2

做了一个功能来完成这项工作：

<?

/*
 * 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;

}

?>

在php中将用户的日期和时间转换为服务器的日期和时间

2 个答案: