What is the correct way to use a variable in a command?

Question

cmd/bat

for /r "%begindir%" /d %%i in (*) do (
  Set "namePath=%%i"
  IF EXIST %%i\test.js (
        set "namePath=!namePath:\=/!"
        set "namePath=!namePath:%begindir%=!"
        set "namePath=!namePath:/test=!"
 		echo !namePath!
 		run -o "{\"tests\": \"\!namePath\!/**/*_test.js\"}"
    )
)

I need to insert it into the line with the launch

run -o "{\"tests\": \"\!namePath\!/**/*_test.js\"}"

. With the addition of this line, I get !namePath! not the correct path. Without a launch line, echo displays the desired values. I did not understand how to correctly escape the variable here.
This is a continuation of my last question ( https://toster.ru/q/507787).

Answer 1

run -o "{\"tests\": \"\!namePath!/**/*_test.js\"}"