通过Unix时间戳的System.DateTime实例的往返转换关闭1小时

注意:为了方便起见,PowerShell用于演示行为,但问题是 System.DateTime System.DateTimeOffset .

这种行为可能有一个很好的概念上的原因,但我没有想到。 如果有的话,了解为什么以及如何避免这种陷阱将是有帮助的。

DateTime 实例通过其Unix时间等价物以本地时间表示:

# Get midnight 1 Jul 2018 in local time.
$date = Get-Date '2018-07-01'

# Convert to Unix time (seconds since midnight 1 Jan 1970 UTC)
# Note: In PowerShell Core this command could be simplified to: Get-Date -Uformat %s $date
$unixTime = [int] (Get-Date -Uformat %s $date.ToUniversalTime())

# Reconvert the Unix time stamp to a local [datetime] instance.
# Note that even though the input string is a UTC time, the cast creates
# a *local* System.DateTime instance (.Kind equals Local)
$dateFromUnixTime1 = ([datetime] '1970-01-01Z').AddSeconds($unixTime)

# Reconvert the Unix time stamp to a local [datetime] instance via 
# a [System.DateTimeOffset] instance:
$dateFromUnixTime2 = ([datetimeoffset ] '1970-01-01Z').AddSeconds($unixTime).LocalDateTime

# Output the results
@"
original:                           $date
Unix time:                          $unixTime
reconstructed via [datetime]:       $dateFromUnixTime1
reconstructed via [datetimeoffset]: $dateFromUnixTime2
"@

以上(关于我的美国英语系统 Eastern Timezone ):

original:                           07/01/2018 00:00:00
Unix time:                          1530417600
reconstructed via [datetime]:       06/30/2018 23:00:00
reconstructed via [datetimeoffset]: 07/01/2018 00:00:00

如你所见 [datetime] 通过 ([datetime] '1970-01-01Z') 实例-其 .Kind 价值是 Local ,即 地方的 日期-关闭1小时,而 [datetimeoffset]

我怀疑这和夏令时有关-不会发生在 2018-12-01