Get the exit value of batch file using vbscript

Question

Please lend some help, I'm a beginner with batch file and vbscript. I would be grateful if you could help me solve my problem.

Given:

a:

@echo off
pushd "C:\Program Files\appName" && popd
IF ERRORLEVEL 1 EXIT 0
pushd "C:\Program Files\appName"
IF EXIST application.exe GOTO Installed
popd
EXIT 1
:Installed
EXIT 2

b:

Set objFSO = CreateObject("Scripting.FileSystemObject")
'Set objFile = objFSO.OpenTextFile(listFile, 1)

dim sh , res

objName = objFSO.GetTempName
objTempFile = objName
'---- Sample Code 0 ----------------------'
'set sh = WScript.CreateObject("WScript.shell")
'res = sh.run("fileFullPath",0,true)
'wScript.echo  res
'----Sample Code 1 ---------------'
'set sh = WScript.CreateObject("WScript.shell")
'res = sh.run("fileName",0,true)
'wScript.echo  res
'---- Sample Code 2 ---------------------
'set sh = WScript.CreateObject("WScript.shell")
'res = sh.run("cmd /c fileName",0,true)
'wScript.echo  res
'---- Sample Code 3 ---------------------
'set sh = WScript.CreateObject("WScript.shell")
'res = sh.run("cmd /c fileFullPath",0,true)
'wScript.echo  res

The b:sample code 1 & 2 works fine with me but I need the sample code 0 & 3. with sample code 1 I always get an error : The system cannot find the file selected ,while with sample code 3 it would only return 0 when failed then 1 when successful, and not the returned value of the batch file.

Thanks,

Sample code 3 worked just fine when I tested it. However, exit code 0 normally means "success", so you shouldn't return it in case of an error. I would also recommend using exit /b instead of just exit. — Ansgar Wiechers
– Ansgar Wiechers, Commented Mar 6, 2014 at 12:45
Thanks for the response @AnsgarWiechers. I already have tried using exit /b before, but it still won't work. Considering that the file I'm going to search does exists. With sample code 1 & 2 it gives the value of 2. While with Sample Code 3, specified the full path of the batch file I keep on having 1 as my output. Is there another way around for me to have same output as sample code 1 & 2 given that I have to access the batch file by it's full path. Thanks. — JSun
– JSun, Commented Mar 6, 2014 at 23:04
Code 1 & 2 should only work when the batch script is in the current working directory. Other than that all 4 samples should yield the same result. Please provide evidence. — Ansgar Wiechers
– Ansgar Wiechers, Commented Mar 7, 2014 at 8:10
Hello @Ansgar-Wiechers. Yes, code 1 & 2 does only works that way. That's why I'm needing 0 & 3. It didn't yield same result because 0 & 3 result is equals to 1 means fail, Thanks for kick --> "0 normally means success". — JSun
– JSun, Commented Mar 7, 2014 at 9:57

JSun · Accepted Answer · 2014-03-07 09:27:22Z

1

I already got my problem solved just a while ago. I found out that even-though i already have had "cmd /c fileFullPath" double-quoted. When a folder name(s) consists some spaces, it would cause an issue --> file can't be found. Now I'm using "cmd /c ""fileFullPath""" instead.

Sorry for having my first post lack of information. I will do my best to make it clearer next time. Thanks a lot. :D

answered Mar 7, 2014 at 9:27

JSun

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