I'm sure this must be possible, but I can't find out how to do it.
Any clues?
I'm sure this must be possible, but I can't find out how to do it.
Any clues?
Use:
ii .
which is short for
Invoke-Item .
The dot can be substituted with any path.
start .
–
Valenti You have a few options:
Examples:
PS C:\> explorer
PS C:\> explorer .
PS C:\> explorer /n
PS C:\> Invoke-Item c:\path\
PS C:\> ii c:\path\
PS C:\> Invoke-Item c:\windows\explorer.exe
PS C:\> ii c:\windows\explorer.exe
PS C:\> [diagnostics.process]::start("explorer.exe")
ii
is an alias of Invoke-Item
–
Eastwood Use any of these:
start .
explorer .
start explorer .
ii .
invoke-item .
You may apply any of these commands in PowerShell.
Just in case you want to open the explorer from the command prompt, the last two commands don't work, and the first three work fine.
Just use the Invoke-Item
cmdlet. For example, if you want to open a explorer window on the current directory you can do:
Invoke-Item .
explorer .
explorer
seems to be out my path –
Mastiff I came across this question looking for a way to open an Explorer window from PowerShell and also select a file. I'm adding this answer in case others come across it for the same reason.
To launch Explorer and select a file, use Invoke-Expression:
Invoke-Expression "explorer '/select,$filePath'"
There are probably other ways to do this, but this worked for me.
Invoke-Expression "explorer '/select,""$filePath""'
–
Haematozoon Invoke-Expression "explorer '/select,""$filePath""'"
–
Hassiehassin $startinfo = new-object System.Diagnostics.ProcessStartInfo
$startinfo.FileName = "explorer.exe"
$startinfo.WorkingDirectory = 'D:\foldername'
[System.Diagnostics.Process]::Start($startinfo)
Hope this helps
start explorer.exe
Simple single line command
This is the only thing that fit my unique constraints of wanting the folder to open as a Quizo Tab in any existing Explorer window.
$objShell = New-Object -ComObject "Shell.Application"
$objShell.Explore("path")
I wanted to write this as a comment but I do not have 50 reputation.
All of the answers in this thread are essentially to use Invoke-Item or to use explorer.exe directly; however, this isn't completely synonymous with "open containing folder", so in terms of opening an Explorer window as the question states, if we wanted to apply the answer to a particular file the question still hasn't really been answered.
e.g.,
Invoke-Item C:\Users\Foo\bar.txt
explorer.exe C:\Users\Foo\bar.html
^ those two commands would result in Notepad.exe or Firefox.exe being invoked on the two files respectively, not an explorer.exe window on C:\Users\Foo\ (the containing directory).
Whereas if one was issuing this command from powershell, this would be no big deal (less typing anyway), if one is scripting and needs to "open containing folder" on a variable, it becomes a matter of string matching to extract the directory from the full path to the file.
Is there no simple command "Open-Containing-Folder" such that a variable could be substituted?
e.g.,
$foo = "C:\Users\Foo\foo.txt"
[some code] $fooPath
# opens C:\Users\Foo\ and not the default program for .txt file extension
for powershell maybe this is an idea for windows 11, i did not figure out how to open all locations/folders in one window with n tabs. $c = @('c:\users', 'c:', 'c:\temp'); Invoke-Item $c
If you need certain credential to access to a folder (or a UNC path), explorer $fooPath will prompt you for the credential, but Invoke-Item or Set-Location $fooPath will throw an ItemNotFoundException.
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ii
isexplorer
, second param is path. – Circuit