How do i handle ERRORLEVEL 9 in cmd

Question

I have developed a batch script which checks for the version of the software installed on a machine and based on the version it picks up the correct version patch and installs it (runs) on the machine.

However i am facing an issue where when i check for a particular string in a file path and if the file doesnt exist then it should go to else but in my case it is still continuing with ERRORLEVEL 1 option. Please have a look at the command below:

sfk find  "%PATH%\Env\Met\Env.xml" "<Version>806</Version>"
Pause
ECHO %ERRORLEVEL%
Pause
IF ERRORLEVEL 1 (GOTO CHECK_RFA_v8.6) ELSE GOTO CHECK_MET_v8.6

Its possible that some clients based on their subscription may have restricted access to the market and hence few markets could be missing in the software. Here in this case as one of the market is missing, the file Env.xml doesnt exist in the path and so i want it to go to else (CHECK_MET_v8.6) rather than proceed with errorlevel 1 (CHECK_RFA_8.6).

What i dont understand is why is it going to CHECK_RFA_v8.6 if the errorlevel is not 1. I have added ECHO %ERRORLEVEL% to check what is it returning and it is 9.

Output:

Where is software installed?
error: no such file or dir: C:\AR\Env\Met\Env.xml
1 errors occurred.
Press any key to continue . . .
9
Press any key to continue . . .
CHECKING RFA v8.6

Can somebody please explain why would it continue with ERRORLEVEL 1 option if the returned error code is 9? and how can i handle it?

Answer 1

If you read the description of if command (available via if /?), you will realize that if errorlevel number command will execute the command when the errorlevel is greater or equal than the number. If you want to execute the command when the number is precisely a certain value, use this form:

IF %ERRORLEVEL% EQU 1 (GOTO CHECK_RFA_v8.6) ELSE GOTO CHECK_MET_v8.6