我尝试过很多不同的方法,但我似乎做错了。
这是我已经尝试过的代码.
[String]$dateValue = '20161212'
[String]$dateStamp = $dateValue -f (Get-Date)
[String]$dateStamp2 = ([datetime]::parseexact($dateValue, "yyyyMMdd", [System.Globalization.CultureInfo]::InvariantCulture)).Date
[String]$dateStamp3 = ([datetime]::FromFileTime($dateValue)).ToString('g')
Write-Host '$dateStamp = ' $dateStamp
Write-Host '$dateStamp2 = ' $dateStamp2
Write-Host '$dateStamp3 = ' $dateStamp3电流码输出
$dateStamp = 20161212
$dateStamp2 = 12/12/2016 00:00:00
$dateStamp3 = 12/31/1600 5:00 PM期望代码输出
$dateStamp = 12/12/2016有什么想法吗?
发布于 2016-04-12 01:58:54
一旦有了datetime对象,就很容易将其转换为所需的任何字符串格式。你离你的第二次尝试太近了。添加ToString允许您指定字符串格式。
([datetime]::parseexact($dateValue, "yyyyMMdd", [System.Globalization.CultureInfo]::InvariantCulture)).ToString("dd/MM/yyyy")https://stackoverflow.com/questions/36562124
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