How do I open a certain file location in PowerShell?

Question

Using Windows PowerShell, is there a shorter way of opening a program other than the following?

cd c:\program files
cd program folder
start program.exe

Answer 1

You can use the Invoke-Item cmdlet and specify the full path as opposed to changing location multiple times to get to the directory containing your target executable

Invoke-Item "C:\Program Files\Program Folder\Program.exe"

Answer 2

You could skip the cd steps by specifying the whole path, e.g c:\path\to\program.exe

If it's a program you launch with some regularity, you could also create an alias. With an alias, you could have PowerShell launch c:\path\to\program.exe with just myapp, for example.

To create an alias, use the following command:

Set-Alias MyApp "C:\Path\To\Program.exe"

Finally, you could also add the folder location to your Windows PATH environment variable. Here's a link to a post detailing how to do this for Python, but you should be able to follow the same instructions for your folder. https://stackoverflow.com/questions/6318156/adding-python-path-on-windows-7