Batch How to start a program - batch-file

Question

I want to create a batch file to launch my executable file after it has made some changes to itself.
My batch file is:
START /D "C:\Users\me\AppData\Roaming\Test\Test.exe"
When I run it though I just get a brief console flash and Test.exe doesn't start up.
I've verified the EXE is there in the directory.
I've launched the exe manually to verify it is working as well.
My batch file resides in
C:\Users\admin\AppData\Roaming\run.bat"

Answer 1

There are two issues:
The /D option solely defines the starting or working directory, but not the program to execute.
The start command considers the first quoted argument as the title of the new window. To avoid confusion with other arguments, always provide a window title (that may also be empty).
There are two solutions, which are actually not exactly equivalent:
Remove the /D option, so the current working directory is used:
start "" "C:\Users\me\AppData\Roaming\Test\Test.exe"
Keep the /D option and explicitly provide the new working directory to be used:
start "" /D "C:\Users\me\AppData\Roaming\Test" "Test.exe"

Answer 2

try changing to this
start /d "C:\Users\me\AppData\Roaming\Test" Test.exe
You will see the console flash and your program should startup.
Update
Thanks for #SomethingDark 's suggestion to use the following code.
start "" C:\Users\me\AppData\Roaming\Test\Test.exe
However, the above code will not work if your filename contains space.

Answer 3

Try with the following command.Add it to your batch script.Notice that you have to add double quotes after start keyword if there is/are whitespaces in the path string.
start "" "C:\Users\me\AppData\Roaming\Test\Test.exe"

Answer 4

Enclose any directory names which are longer than one-word into quotation marks. So the following path:
start C:\Program Files\MySQL\MySQL Workbench 8.0 CE\MySQL.exe
Should become something like this:
start C:\"Program Files"\MySQL\"MySQL Workbench 8.0 CE"\MySQL.exe