Get java version from batch file

it will be surly idiot but why do not just print JAVA_HOME path ?

You can do this with awk:

>java -fullversion 2>&1|awk "{print $NF}"
"1.7.0_21-b11"

>java -fullversion 2>&1|awk -F\" "{print $(NF-1)}"
1.7.0_21-b11

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Script example:

@ECHO OFF &SETLOCAL
FOR /f %%a IN ('java -fullversion 2^>^&1^|awk "{print $NF}"') DO SET "javaversion=%%a"
IF DEFINED javaversion (ECHO java version: %javaversion%) ELSE ECHO java NOT found

output is: java version: "1.7.0_21-b11"
awk for Windows

for /f tokens^=2-5^ delims^=.-_^" %j in ('java -fullversion 2^>^&1') do @set "jver=%j%k%l%m"

This will store the java version into jver variable and as integer And you can use it for comparisons .E.G

if %jver% LSS 16000 echo not supported version

.You can use more major version by removing %k and %l and %m.This command prompt version.

For .bat use this:

@echo off
PATH %PATH%;%JAVA_HOME%\bin\
for /f tokens^=2-5^ delims^=.-_^" %%j in ('java -fullversion 2^>^&1') do set "jver=%%j%%k%%l%%m"

According to my tests this is the fastest way to get the java version from bat (as it uses only internal commands and not external ones as FIND,FINDSTR and does not use GOTO which also can slow the script). Some JDK vendors does not support -fullversion switch or their implementation is not the same as this one provided by Oracle (better avoid them).