5

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.

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5
  • Last time I checked java installer put java.exe in c:\windows\system32\java.exe. So are you sure you really want to do that? Commented Jun 1, 2010 at 17:27
  • On my machine JRE installs into: c:\Program Files\Java\jre6\bin\java.exe Commented Jun 1, 2010 at 17:32
  • Note that JAVA_HOME is only used by the JDK. Its absence doesn't tell you anything about whether a JRE is installed. And java.exe is part of both ... Commented Jun 1, 2010 at 17:49
  • Ignore JAVA_HOME in my code example. It could be my own env. variable which I want to setup from batch script for future use. Commented Jun 1, 2010 at 18:18
  • Ok. this is what I have got so far: <code> for /f %%j in ("java.exe") do ( set REAL_JAVA_PATH="%%~dp$PATH:j" ) </code> Commented Jun 1, 2010 at 18:19

4 Answers 4

Reset to default
6

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.

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1 Comment

I installed JRE7-32bit on Windows 8. reg.exe does not recognize any version of Java. I can run executable jars, but I can't use any java command in the console like java or javaw. How can I detect if Java is installed in such a system?
4

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).
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1 Comment

On my machine "where java" produces a 32 bit version of java while " java -version" finds a 64 bit version.
1

Most recent versions write to the registry:

HKEY_LOCAL_MACHINE\SOFTWARE\JavaSoft

You can look for what keys are in there and get out the path using reg.exe

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2 Comments

Can I be sure that all java versions on all windows versions will have the same path in registry?
This reg path is probably pretty standard, I just don't know how long JAVA has been using the same path(s)/keys. If you wanted to be absolutely sure, you could distribute the java exe with whatever app you're trying to run.
1

I would (using batch)

::get javaw.exe from the latest properly installed jre
for /f tokens^=2^ delims^=^" %%i in ('reg query HKEY_CLASSES_ROOT\jarfile\shell\open\command /ve') do set JAVAW_PATH=%%i

::if reg entry is not found, java is not installed
if "%JAVAW_PATH%"=="" goto YOUR_ERROR

::then strip "\javaw.exe" from the JAVAW_PATH obtained above
set JAVA_HOME=%JAVAW_PATH:\javaw.exe=%

This should work with jre 1.6 and 1.7 installed in windows xp and seven and be way more faster than searching the filesystem.

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