Batch file to check the minimum required Java version

Question

I'm new to batch script coding and a little stuck here. I've got a .sh file that checks if version of java installed is at least 1.8.0:

# Minimal version
MINIMAL_VERSION=1.8.0

# Check if Java is present and the minimal version requirement
_java=`type java | awk '{ print $ NF }'`
CURRENT_VERSION=`"$_java" -version 2>&1 | awk -F'"' '/version/ {print $2}'` #says 1.8.0_65
minimal_version=`echo $MINIMAL_VERSION | awk -F'.' '{ print $2 }'` #says 8
current_version=`echo $CURRENT_VERSION | awk -F'.' '{ print $2 }'` #says 8

if [ $current_version ]; then
  if [ $current_version -lt $minimal_version ]; then
    echo "Error: Java version is too low. At least Java >= ${MINIMAL_VERSION} needed.";
    exit 1;
  fi
    else
      echo "Not able to find Java executable or version. Please check your Java installation.";
      exit 1;
fi

It is pretty clear. I need to write a .bat file for Windows exactly same logic. So here's where I'm stuck, because I do not know windows analogs for awk.

Answer 1

Here you can see how you can get java version like integer which can numerically compared:

@echo off
:: uncomment the line bellow if the java.exe is not in the %PATH% 
::PATH %PATH%;%JAVA_HOME%\bin\

java -version 1>nul 2>nul || (
   echo no java installed
   exit /b 2
)
for /f tokens^=2-5^ delims^=.-_^" %%j in ('java -fullversion 2^>^&1') do set "jver=%%j%%k%%l%%m"

if %jver% LSS 18000 (
  echo java version is too low 
  echo at least 1.8 is needed
  exit /b 1
)