Как найти java sdk

I need to know where JDK is located on my machine.

On running Java -version in cmd, it shows the version as ‘1.6.xx’.
To find the location of this SDK on my machine I tried using echo %JAVA_HOME% but it is only showing ‘JAVA_HOME’ (as there is no ‘JAVA_PATH’ var set in my environment variables).

In windows the default is: C:Program FilesJavajdk1.6.0_14 (where the numbers may differ, as they’re the version).

Java installer puts several files into %WinDir%System32 folder (java.exe, javaws.exe and some others). When you type java.exe in command line or create process without full path, Windows runs these as last resort if they are missing in %PATH% folders.

You can lookup all versions of Java installed in registry. Take a look at HKLMSOFTWAREJavaSoftJava Runtime Environment and HKLMSOFTWAREWow6432NodeJavaSoftJava Runtime Environment for 32-bit java on 64 bit Windows.

This is how java itself finds out different versions installed. And this is why both 32-bit and 64-bit version can co-exist and works fine without interfering.

answered Feb 1, 2012 at 10:48

The batch script below will print out the existing default JRE. It can be easily modified to find the JDK version installed by replacing the Java Runtime Environment with Java Development Kit.

@echo off

setlocal

::- Get the Java Version
set KEY="HKLMSOFTWAREJavaSoftJava Runtime Environment"
set VALUE=CurrentVersion
reg query %KEY% /v %VALUE% 2>nul || (
    echo JRE not installed 
    exit /b 1
)
set JRE_VERSION=
for /f "tokens=2,*" %%a in ('reg query %KEY% /v %VALUE% ^| findstr %VALUE%') do (
    set JRE_VERSION=%%b
)

echo JRE VERSION: %JRE_VERSION%

::- Get the JavaHome
set KEY="HKLMSOFTWAREJavaSoftJava Runtime Environment%JRE_VERSION%"
set VALUE=JavaHome
reg query %KEY% /v %VALUE% 2>nul || (
    echo JavaHome not installed
    exit /b 1
)

set JAVAHOME=
for /f "tokens=2,*" %%a in ('reg query %KEY% /v %VALUE% ^| findstr %VALUE%') do (
    set JAVAHOME=%%b
)

echo JavaHome: %JAVAHOME%

endlocal

answered Aug 11, 2014 at 18:21

In Windows PowerShell you can use the Get-Command function to see where Java is installed:

Get-Command -All java

Or

gcm -All java

The -All part makes sure to show all places it appears in the Path lookup. Below is example output.

PS C:> gcm -All java

CommandType     Name                                               Version    Source
-----------     ----                                               -------    ------
Application     java.exe                                           8.0.202.8  C:Program Files (x86)Common FilesOracleJavajav...
Application     java.exe                                           8.0.131... C:ProgramDataOracleJavajavapathjava.exe

answered Nov 9, 2020 at 17:32

More on Windows… variable java.home is not always the same location as the binary that is run.

As Denis The Menace says, the installer puts Java files into Program Files, but also java.exe into System32. With nothing Java related on the path java -version can still work. However when PeterMmm’s program is run it reports the value of Program Files as java.home, this is not wrong (Java is installed there) but the actual binary being run is located in System32.

One way to hunt down the location of the java.exe binary, add the following line to PeterMmm’s code to keep the program running a while longer:

try{Thread.sleep(60000);}catch(Exception e) {}

Compile and run it, then hunt down the location of the java.exe image. E.g. in Windows 7 open the task manager, find the java.exe entry, right click and select ‘open file location’, this opens the exact location of the Java binary. In this case it would be System32.

answered Oct 24, 2012 at 10:23

Have you tried looking at your %PATH% variable. That’s what Windows uses to find any executable.

On macOS, run:

cd /tmp && echo 'public class Main {public static void main(String[] args) {System.out.println(System.getProperty("java.home"));}}' > Main.java && javac Main.java && java Main

On my machine, this prints:

/Library/Java/JavaVirtualMachines/jdk-9.0.4.jdk/Contents/Home

Note that running which java does not show the JDK location, because the java command is instead part of JavaVM.framework, which wraps the real JDK:

$ which java
/usr/bin/java
/private/tmp
$ ls -l /usr/bin/java
lrwxr-xr-x  1 root  wheel  74 14 Nov 17:37 /usr/bin/java -> /System/Library/Frameworks/JavaVM.framework/Versions/Current/Commands/java

answered Jan 18, 2018 at 13:58

None of these answers are correct for Linux if you are looking for the home that includes the subdirs such as: bin, docs, include, jre, lib, etc.

On Ubuntu for openjdk1.8.0, this is in:
/usr/lib/jvm/java-1.8.0-openjdk-amd64

and you may prefer to use that for JAVA_HOME since you will be able to find headers if you build JNI source files. While it’s true which java will provide the binary path, it is not the true JDK home.

answered Jan 27, 2016 at 1:33

I have improved munsingh’s answer above by testing for the registry key in 64-bit and 32-bit registries, if needed:

::- Test for the registry location  
SET VALUE=CurrentVersion
SET KEY_1="HKLMSOFTWAREJavaSoftJava Development Kit"
SET KEY_2=HKLMSOFTWAREJavaSoftJDK
SET REG_1=reg.exe
SET REG_2="C:Windowssysnativereg.exe"
SET REG_3="C:Windowssyswow64reg.exe"

SET KEY=%KEY_1%
SET REG=%REG_1%
%REG% QUERY %KEY% /v %VALUE% 2>nul
IF %ERRORLEVEL% EQU 0 GOTO _set_value

SET KEY=%KEY_2%
SET REG=%REG_1%
%REG% QUERY %KEY% /v %VALUE% 2>nul
IF %ERRORLEVEL% EQU 0 GOTO _set_value

::- %REG_2% is for 64-bit installations, using "C:Windowssysnative"
SET KEY=%KEY_1%
SET REG=%REG_2%
%REG% QUERY %KEY% /v %VALUE% 2>nul
IF %ERRORLEVEL% EQU 0 GOTO _set_value

SET KEY=%KEY_2%
SET REG=%REG_2%
%REG% QUERY %KEY% /v %VALUE% 2>nul
IF %ERRORLEVEL% EQU 0 GOTO _set_value

::- %REG_3% is for 32-bit installations on a 64-bit system, using "C:Windowssyswow64"
SET KEY=%KEY_1%
SET REG=%REG_3%
%REG% QUERY %KEY% /v %VALUE% 2>nul
IF %ERRORLEVEL% EQU 0 GOTO _set_value

SET KEY=%KEY_2%
SET REG=%REG_3%
%REG% QUERY %KEY% /v %VALUE% 2>nul
IF %ERRORLEVEL% EQU 0 GOTO _set_value

:_set_value
FOR /F "tokens=2,*" %%a IN ('%REG% QUERY %KEY% /v %VALUE%') DO (
    SET JDK_VERSION=%%b
)
SET KEY=%KEY%%JDK_VERSION%
SET VALUE=JavaHome
FOR /F "tokens=2,*" %%a IN ('%REG% QUERY %KEY% /v %VALUE%') DO (
    SET JAVAHOME=%%b
)
ECHO "%JAVAHOME%"
::- SETX JAVA_HOME "%JAVAHOME%"

reference for access to the 64-bit registry

answered Mar 28, 2018 at 17:51

Maybe the above methods work… I tried some and didn’t for me. What did was this :

Run this in terminal :

/usr/libexec/java_home

answered Oct 17, 2020 at 3:13

Simple method (Windows):
Open an application using java.
press ctrl + shift + esc

Right click on OpenJDK platform binary. Click open file location.
Then it will show java/javaw.exe then go to the top where it shows the folder and click on the jdk then right copy the path, boom. (Wont work for apps using bundled jre paths/runtimes, because it will show path to the bundled runtime)

Script for 32/64 bit Windows.

@echo off

setlocal enabledelayedexpansion

::- Get the Java Version
set KEY="HKLMSOFTWAREJavaSoftJava Runtime Environment"
set KEY64="HKLMSOFTWAREWOW6432NodeJavaSoftJava Runtime Environment"
set VALUE=CurrentVersion
reg query %KEY% /v %VALUE% 2>nul || (
    set KEY=!KEY64!
    reg query !KEY! /v %VALUE% 2>nul || (
    echo JRE not installed 
    exit /b 1
)
)

set JRE_VERSION=
for /f "tokens=2,*" %%a in ('reg query %KEY% /v %VALUE% ^| findstr %VALUE%') do (
    set JRE_VERSION=%%b
)

echo JRE VERSION: %JRE_VERSION%

::- Get the JavaHome
set KEY="HKLMSOFTWAREJavaSoftJava Runtime Environment%JRE_VERSION%"
set KEY64="HKLMSOFTWAREWOW6432NodeJavaSoftJava Runtime Environment%JRE_VERSION%"
set VALUE=JavaHome
reg query %KEY% /v %VALUE% 2>nul || (
    set KEY=!KEY64!
    reg query !KEY! /v %VALUE% 2>nul || (
    echo JavaHome not installed
    exit /b 1
)
)

set JAVAHOME=
for /f "tokens=2,*" %%a in ('reg query %KEY% /v %VALUE% ^| findstr %VALUE%') do (
    set JAVAHOME=%%b
)

echo JavaHome: %JAVAHOME%

endlocal

answered Mar 26, 2019 at 23:23