用户的日期和时间转换为服务器的日期和时间在php

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当天和UTC时间。我有每个用户的gmt抵消金额（用户在注册时设置）。例如：

在东部时区的用户选择：

下午3:15，周一，周二，周五

我需要知道什么时间和日期的信息将是UTC时间。解决方案必须考虑到在一个时区中的这种星期一可能是UTC时间的另一天。另外，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，沿着阵列线的东西应返回如：

Array('<3:15 pm eastern adjusted for utc>', '<Monday adjusted for UTC>', '<Tuesday adjusted for UTC>', '<Friday adjusted for UTC>'); 

我不需要结果直接格式化为这样的阵列 - 这仅仅是最终目标。

我猜它涉及使用strtotime，但我只是不能完全指出如何去做。

Answer 1

做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; 

} 

?> 

Answer 2

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

结果将是一个时间戳，这里有一个例子：

$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：总是忘记小事，应该可以完美工作了。