How to get Java Version from batch script?

Question

I am trying to get \'6\' out of the java version output given below

java version \"1.6.0_21\"
Java(TM) SE Runtime Environment (build 1.6.0_21-b07)
Java HotSp         


        

Answer 1

I've made some modification to Patrick's answer so it works with Java 9, 10, etc. This returns minor version for 1.x, and major version for Java 9, 10, etc.

@echo off
setlocal

rem We use the value the JAVACMD environment variable, if defined
rem and then try JAVA_HOME
set "_JAVACMD=%JAVACMD%"
if "%_JAVACMD"=="" (
  if not "%JAVA_HOME%"=="" (
    if exist "%JAVA_HOME%\bin\java.exe" set "_JAVACMD=%JAVA_HOME%\bin\java.exe"
  )
)
if "%_JAVACMD%"=="" set _JAVACMD=java

rem Parses x out of 1.x; for example 8 out of java version 1.8.0_xx
rem Otherwise, parses the major version; 9 out of java version 9-ea
set JAVA_VERSION=0
for /f "tokens=3" %%g in ('%_JAVACMD% -Xms32M -Xmx32M -version 2^>^&1 ^| findstr /i "version"') do (
  set JAVA_VERSION=%%g
)
set JAVA_VERSION=%JAVA_VERSION:"=%
for /f "delims=.-_ tokens=1-2" %%v in ("%JAVA_VERSION%") do (
  if /I "%%v" EQU "1" (
    set JAVA_VERSION=%%w
  ) else (
    set JAVA_VERSION=%%v
  )
)

@echo %JAVA_VERSION%

endlocal