Start windows app via start Menu

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JesperRed
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Start windows app via start Menu

Post by *JesperRed »

I would like to start the Microsoft app "FileViewerFinder" from the Start Menu in TCM
I have found the location for FileViewerFinder to be
C:\Program Files\WindowsApps\D5BE6627.FileViewerFinder_4.1.0.0_x86__9pm2v9747qaaa
(see how https://www.ghisler.ch/board/viewtopic.php?f=3&t=58413&p=371898&hilit=microsoft+apps#p371898)

When in insert the location in the start menu and execute it, then I'm send to
this adress C:\Program Files\WindowsApps\D5BE6627.FileViewerFinder_4.1.0.0_x86__9pm2v9747qaaa in File explorer
The program is not been started.

What am I missing?
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petermad
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Re: Start windows app via start Menu

Post by *petermad »

I don't have the app myself, but try with this as the command:

Code: Select all

%SystemRoot%\explorer.exe shell:appsFolder\D5BE6627.FileViewerFinder_9pm2v9747qaaa!App
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IvoA1975
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Re: Start windows app via start Menu

Post by *IvoA1975 »

I tried above for the Typesy app. I found the location to be:
shell:appsFolder\eReflectPtyLtd.Typesy_1.0.0.0_x64__3v9eb52zjj184\app\Typesy.exe

I tried starting it in the start menu by adding:
"%SystemRoot%\explorer.exe shell:appsFolder\eReflectPtyLtd.Typesy_1.0.0.0_x64__3v9eb52zjj184\app\Typesy.exe!App"
"%SystemRoot%\explorer.exe shell:appsFolder\eReflectPtyLtd.Typesy_1.0.0.0_x64__3v9eb52zjj184\app\Typesy.exe"
"%SystemRoot%\explorer.exe shell:appsFolder\eReflectPtyLtd.Typesy_1.0.0.0_x64__3v9eb52zjj184!App"

None of these worked.
Can anyone explain me how to add Apps? It would be good if I could also start Apps in the menu, instead of only Windows desktop applications.
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Re: Start windows app via start Menu

Post by *Dalai »

2IvoA1975
Try

Code: Select all

"%SystemRoot%\explorer.exe" shell:appsFolder\eReflectPtyLtd.Typesy_3v9eb52zjj184!App
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IvoA1975
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Re: Start windows app via start Menu

Post by *IvoA1975 »

Thanks for the reply, it didn't work it just started the Windows explorer.
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Re: Start windows app via start Menu

Post by *Dalai »

Well, then you may need to find the right app ID and name via the PowerShell Get-AppxPackage command, as I wrote in the thread JesperRed linked to.

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Re: Start windows app via start Menu

Post by *IvoA1975 »

Hi,

It lists the following, but this means the following should work right?
"%SystemRoot%\explorer.exe shell:appsFolder\eReflectPtyLtd.Typesy_1.0.0.0_x64__3v9eb52zjj184!App"

Name : eReflectPtyLtd.Typesy
Publisher : CN=0C7CB0BA-3DF8-4490-9E8F-28F010854FB3
Architecture : X64
ResourceId :
Version : 1.0.0.0
PackageFullName : eReflectPtyLtd.Typesy_1.0.0.0_x64__3v9eb52zjj184
InstallLocation : C:\Program Files\WindowsApps\eReflectPtyLtd.Typesy_1.0.0.0_x64__3v9eb52zjj184
IsFramework : False
PackageFamilyName : eReflectPtyLtd.Typesy_3v9eb52zjj184
PublisherId : 3v9eb52zjj184
IsResourcePackage : False
IsBundle : False
IsDevelopmentMode : False
NonRemovable : False
IsPartiallyStaged : False
SignatureKind : Store
Status : Ok
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Dalai
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Re: Start windows app via start Menu

Post by *Dalai »

IvoA1975 wrote: 2020-07-09, 09:11 UTC[...] but this means the following should work right?
"%SystemRoot%\explorer.exe shell:appsFolder\eReflectPtyLtd.Typesy_1.0.0.0_x64__3v9eb52zjj184!App"
No! First, get rid of the second quote at the end. Either don't use any quotes at all, or only quote the path to explorer.exe. Second, remove any version numbers and architectures from the command. They must not be there. Just the vendor name, app name and app/publisher ID; it's more or less what's given for PackageFamilyName (but I'm not sure if that holds true for every app).

I suggest to inspect

Code: Select all

C:\Program Files\WindowsApps\eReflectPtyLtd.Typesy_1.0.0.0_x64__3v9eb52zjj184\AppxManifest.xml
and look for the following XML tag

Code: Select all

<Application ID="<id_here>" Executable="<some_name>" ...>
You need that ID to be able to start an app. Once you have it, replace the "!App" I posted above with "!<your_ID>" and it should work.

It's the same with the Immersive Control Panel and Windows Security App. Their ID is not "App" but something else, which is not shown in the Get-AppxPackage command's output, unfortunately.

See also https://www.tenforums.com/software-apps/57000-method-open-any-windows-10-apps-command-line.html

-----

Request to any coders with PowerShell knowledge out there:
It would be great if the determination of the app IDs could be made easier, e.g. by interpreting/parsing the AppxManifest.xml files, searching for the <Application ID> tag and extracting the ID. This much manual work to get the apps to launch (from TC, Batch and so on) is really annoying (although, in my opinion, this is also Microsoft's fault - again).

Regards
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Re: Start windows app via start Menu

Post by *Horst.Epp »

In Quick Access PopUp I can also start Windows apps.
The script to get the usable Ids is:

Code: Select all

$installedapps = get-AppxPackage
$ids = $null
foreach ($app in $installedapps)
{
try
{
$ids = (Get-AppxPackageManifest $app -erroraction Stop).package.applications.application.id
}
catch
{
Write-Output "No Id's found for $($app.name)"
}
foreach ($id in $ids)
{
$line = $app.Name + "	" + $app.packagefamilyname + "!" + $id
echo $line
$line >> 'C:\Tools\QuickAccessPopup\_QAP_temp_1158859728\CollectWindowsAppsList.tsv'
}
}
# write-host "Press any key to continue..."
# [void][System.Console]::ReadKey($true)
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Dalai
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Re: Start windows app via start Menu

Post by *Dalai »

2Horst.Epp
Thank you! Didn't know that Get-AppxPackageManifest exists (and I admit I didn't look for such a command :oops:). Anyway, the script is really useful. Is there a way to make the output look nicer, like in columns (as the column command does on Linux)?

Regards
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Re: Start windows app via start Menu

Post by *Dalai »

OK, it looks like I got it. Here's an updated script based on Horst.Epp's:

Code: Select all

$installedapps = Get-AppxPackage
$ids = $null
$max=0

foreach ($app in $installedapps)
{
  if ($app.Name.Length -gt $max)
  {
    $max = $app.Name.Length
  }
}
foreach ($app in $installedapps)
{
    try
    {
        $ids = (Get-AppxPackageManifest $app -ErrorAction Stop).Package.Applications.Application.ID
    }
    catch
    {
        Write-Output "No IDs found for $($app.Name)"
    }
    foreach ($id in $ids)
    {
        "{0,-$max}  :   {1}!{2}" -f $app.Name,$app.PackageFamilyName,$id
    }
}
# write-host "Press any key to continue..."
# [void][System.Console]::ReadKey($true)
Regards
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IvoA1975
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Re: Start windows app via start Menu

Post by *IvoA1975 »

Hi,
The tag is the following one:
<Application Id="Typesy" Executable="app\Typesy.exe" EntryPoint="Windows.FullTrustApplication">

So I tried the following (but it did not work):

%SystemRoot%\explorer.exe shell:appsFolder\eReflectPtyLtd.Typesy!App
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Horst.Epp
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Re: Start windows app via start Menu

Post by *Horst.Epp »

Dalai wrote: 2020-07-09, 12:26 UTC 2Horst.Epp
Thank you! Didn't know that Get-AppxPackageManifest exists (and I admit I didn't look for such a command :oops:). Anyway, the script is really useful. Is there a way to make the output look nicer, like in columns (as the column command does on Linux)?

Regards
Dalai
The Author of this script is the author of Quick Access PopUp Jean Lalonde.
May be he can help how to start such apps succesfully because it works in QAP :D
QAP is written in Autohotkey.
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Dalai
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Re: Start windows app via start Menu

Post by *Dalai »

2IvoA1975

Code: Select all

"%SystemRoot%\explorer.exe" shell:appsFolder\eReflectPtyLtd.Typesy_3v9eb52zjj184!Typesy
should work.

You can also use the script Horst.Epp posted above (or my adaptation for a nicer output) to find the correct app ID. Save it as ListAppsWithIds.ps1 in a directory of your choice and run it like this:

Code: Select all

powershell.exe -ExecutionPolicy RemoteSigned -File d:\rive\path\ListAppsWithIds.ps1
Regards
Dalai
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Re: Start windows app via start Menu

Post by *IvoA1975 »

Great, thnx, that worked!

So for myself I made the following procedure on how to add an App to the Total Commander menu, I will share it here below:
(I tested this procedure with a couple of Apps and they all worked)


First find the [PackageFamilyName] by:
* starting Windows Powershell
* Type command: Get-AppxPackage

Search in the output for the App and note PackageFamilyName
* Example for Typesy: eReflectPtyLtd.Typesy_3v9eb52zjj184

Next, find the app location and go into the AppxManifest.xml
* via shell:appsFolder
* or via C:\Program Files\WindowsApps (hidden system directory)
* Get the [AppID] by searching in de AppxManifest.xml XML file for the tag <Application Id=...>

Then start the App with the following command:
"%SystemRoot%\explorer.exe" shell:appsFolder\[PackageFamilyName]![AppID]

Example for Typesy: "%SystemRoot%\explorer.exe" shell:appsFolder\eReflectPtyLtd.Typesy_3v9eb52zjj184!Typesy

The ICON file if the exe file in the App folder
* Example for Typesy: "C:\Program Files\WindowsApps\eReflectPtyLtd.Typesy_1.0.0.0_x64__3v9eb52zjj184\app\Typesy.exe"
* But it's easier to just make a Snagit of an image search and save it as ICO file
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