Question

下午好，

有人可以告诉我如何使用powershell将日期20120624转换为yyyyMMdd，因为我得到了＃34;字符串未被识别为valud日期时间＆＃34;运行以下代码时

[String]$ServiceTag = "b26ybt1";

Try{ 
       $AssetService = New-WebServiceProxy -Uri "http://xserv.dell.com/services/AssetService.asmx?WSDL"; 
       $ApplicationName = "AssetService"; 
       $Guid = [Guid]::NewGuid(); 
       $Asset = $AssetService.GetAssetInformation($Guid,$ApplicationName,$ServiceTag); 
       $Writer = New-Object "System.IO.StringWriter"; 
       $XmlSerializer = New-Object System.Xml.Serialization.XmlSerializer($Asset.GetType()); 
       $XmlSerializer.Serialize($Writer,$Asset); 
       [String]$Result = $Writer.ToString(); 
       $Writer.Flush(); 
       $Writer.Close(); 
       Return $Result; 
} 
Catch{ 
       Write-Host $($_.Exception.Message);     
}

$prog = [regex]::match($Result,'(?<=StartDate>)(.*)(?=T00)').Groups[1].Value
[System.Text.RegularExpressions.Regex]::Replace($prog,"[-]","");

[datetime]::ParseExact($prog,"yyyyMMdd",$null)

Answer 1

更改

[System.Text.RegularExpressions.Regex]::Replace($prog,"[-]","");

到

$prog = [System.Text.RegularExpressions.Regex]::Replace($prog,"[-]","");

您所做的只是返回替换所做的事情，您从未将其实际分配给变量。当我按上述方式运行时，我得到了以下内容：

Sunday, June 24, 2012 12:00:00 AM

Answer 2

您可以将结果转换为XML而不是String，并避免字符串解析：

[xml]$Result = $Writer.ToString()
[datetime]$Result.ArrayOfAsset.Asset.Entitlements.EntitlementData[0].StartDate

Powershell ParseExact日期格式化yyyyMMdd问题

2 个答案: