How to check Java installation from batch script?

Question

I need to write a batch script to find out if Java is installed, and if it is, then under what path? I feel it has to be something similar to this:

for /f %%j in ("java.exe") do (
   set JAVA_HOME=..........
)

but I can not figure it out.

P.S. It has to work with path with spaces two. Like if java is installed into "Program Files".

Thanks.

Answer 1

Using reg[.exe] you can query possible JRE candidates that are installed on the system. There could be none or could be several.

On a test set-up, running inside the command shell:

reg query "HKLM\Software\JavaSoft\Java Runtime Environment"

I get three result lines, of which the first is CurrentVersion REG_SZ 1.6

Based on that, querying

reg query "HKLM\Software\JavaSoft\Java Runtime Environment\1.6\"

Gives me JavaHome REG_SZ C:\Program Files\Java\jre6

It's much more efficient than scan a file system to find a java binary.

This was tested under a virtual installation of Windows XP 32-bit.

Answer 2

Couldn't you use the 'where' command? As in:

>where java

And test against this?

Example:

C:\Users\myname>where java
C:\Program Files (x86)\Java\jdk1.6.0_17\bin\java.exe

C:\Users\myname>where foo
INFO: Could not find files for the given pattern(s).