Set Java path in command line for only one directory

Question

I am running a program that utilizes Scala 2.10 for work and is not compatible with Java 8, only Java 7. In a Windows 7 command line, how can I set the java path to use Java 7 ONLY for that directory?

Answer 1

If the program uses a batch to start, then add this line before the start of the program:

SET JAVA_HOME="C:\Program Files\Java7\Java.exe"

(This is just an example, the directory might be different on your computer)

If the program does not use such a batch (you can recognize it because it ends with either .cmd or .bat) create such a file and use that for launching the program instead:

@echo off
SET JAVA_HOME=...
ThisIsMyFancyScalaProgram.Exe

Answer 2

You may create 2 batch file one for java 7 and one for java 8 like this -

jdk7.bat

@echo off
echo Setting JAVA_HOME
set JAVA_HOME=C:\Program Files\Java\jdk1.7.0_11
echo setting PATH
set PATH=C:\Program Files\Java\jdk1.7.0_11\bin;%PATH%
echo Display java version
java -version

jdk8.bat

@echo off
echo Setting JAVA_HOME
set JAVA_HOME=C:\Program Files\Java\jdk1.8.0_11
echo setting PATH
set PATH=C:\Program Files\Java\jdk1.7.8_11\bin;%PATH%
echo Display java version
java -version

You may quickly switch between them running these batch file.

Answer 3

To Add system environment variables:

setx JAVA_HOME "C:\Program Files\Java\jdk1.8.0"

setx PATH "%PATH%;%JAVA_HOME%\bin";

To update system environment variables:

setx -m JAVA_HOME "C:\Program Files\Java\jdk1.8.0"

setx -m PATH "%PATH%;%JAVA_HOME%\bin";