nodejs child_process exec'java -version'

Question

hey i'm running the following code in nodejs shell

 exec = require('child_process').exec 

now i have exec var as function. then i run

 exec('java',function (error, stdout, stderr) {console.log(error)});

and

 exec('java -version',function (error, stdout, stderr) {console.log(error)})

the first output is:

  { [Error: Command failed: The system cannot find the path specified.

but the output continue and i get the real output from java ie :

     Usage: java [-options] class [args...]
     (to execute a class)
     or  java [-options] -jar jarfile [args...] 
     (etc....)

the same goes for the my 2nd try with 'java -version' i get

   { [Error: Command failed: The system cannot find the path specified.
     java version "1.7.0_10"
     Java(TM) SE Runtime Environment (build 1.7.0_10-b18)
     Java HotSpot(TM) 64-Bit Server VM (build 23.6-b04, mixed mode)
     ] killed: false, code: 1, signal: null }

any ideas? thanks oak

ps

i'm running on Windows 8 i have JAVA_HOME set and i have %JAVA_HOME%\\bin in my PATH if i run java -version from command line everything is ok. "node -version" v0.10.22

EDIT:

from nodejs doc

"On success, error will be null. On error, error will be an instance of Error and err.code will be the exit code of the child process, and err.signal will be set to the signal that terminated the process."

Answer 1

funny, after investigating nodejs source code i saw exec function use 'cmd /s /c [command]' then i try to run the same command without cmd. found out that i still get this message . turns out that this bug is related to cmd so i searched for it and came up with solution.

for the right answer go here: https://stackoverflow.com/a/13343337/1211174

summary the answer: cmd has option of autorun. and this auto run faild. then cmd write to stderr and then many nodejs faild - because they just check for if (error != null) and not check for error.code ...