Powershell find latest file and open in file explorer

user_14529561 0
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This is based on another question: https://stackoverflow.com/a/9675811/14529561.

How would I pass the result from gci path | sort LastWriteTime | select -last 1 and open with explorer.exe?

I have tried:

$wd = gci path | sort LastWriteTime | select -last 1;
explorer $wd

gci path | sort LastWriteTime | select -last 1 | Format-Table -hidetableheaders | explorer $_.

gci path | sort LastWriteTime | select -last 1 | Format-Table -hidetableheaders | ii $_.

Unfortunately, all the above give me give me errors.

powershell scripting
2 Answers

You can reference the PSParentPath property.

Get-ChildItem -Path path |
    Sort-Object -Property LastWriteTime |
        Select-Object -Last 1 | Foreach-Object {
                Invoke-Item $_.psparentpath
        }
user_40853310

You can do it like this :

$Mypath=$Env:Temp
gci $Mypath | sort LastWriteTime | select -last 1 | % { Explorer /n,/select,$_.FullName }

And if you want to explore in more than folder, you can put thel into an array and do like this :

cls
$Mypath=@("$Env:UserProfile\desktop","$Env:Temp","$Env:AppData")

ForEach ($P in $Mypath) {
    gci $P | sort LastWriteTime | select -last 1 | % { Explorer /n,/select,$_.FullName }
}
user_30807700
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