How to get Java Version from batch script? - java

I am trying to get '6' out of the java version output given below
java version "1.6.0_21"
Java(TM) SE Runtime Environment (build 1.6.0_21-b07)
Java HotSpot(TM) Client VM (build 17.0-b17, mixed mode, sharing)
For the same I wrote this batch script
set VERSION6="1.6.0_21"
java -version 2>&1 | findstr "version" >ab.txt
for /f "tokens=3" %%g in (ab.txt) do (
if not %%g == %VERSION6% echo %%g
echo %%g
)
%%g displays "1.6.0_21"
May someone guide me to correct direction? I am not much familiar with for /f.


#echo off
setlocal
set VERSION6="1.6.0_21"
for /f "tokens=3" %%g in ('java -version 2^>^&1 ^| findstr /i "version"') do (
#echo Output: %%g
set JAVAVER=%%g
)
set JAVAVER=%JAVAVER:"=%
#echo Output: %JAVAVER%
for /f "delims=. tokens=1-3" %%v in ("%JAVAVER%") do (
#echo Major: %%v
#echo Minor: %%w
#echo Build: %%x
)
endlocal
In the first for loop, "tokens=3" says that we're going to just use the third token from the command output. Rather than redirect the output of the java -version command to a file, we can run this command within the for loop itself. The carets (^) are escape characters, and are needed so we can embed the >, & and | symbols in the command string.
Within the body of the for loop, we set a new var, JAVAVER, so that we can do some manipulation of the version string later.
The set JAVAVER=%JAVAVER:"=% command removes the double quotes from around the version string.
The last for loop parses the java version string. delims=. says we're going to delimit tokens using periods. tokens=1-3 says we're going to pass the first three tokens from the string to the body of the loop. We can now get the components of the java version string using the explicit variable, %%v and the implied variables (next letters in the alphabet) %%w and %%x.
When I run this on my system I get:
Output: "1.6.0_24"
Output: 1.6.0_24
Major: 1
Minor: 6
Build: 0_24


I've made some modification to Patrick's answer so it works with Java 9, 10, etc.
This returns minor version for 1.x, and major version for Java 9, 10, etc.
#echo off
setlocal
rem We use the value the JAVACMD environment variable, if defined
rem and then try JAVA_HOME
set "_JAVACMD=%JAVACMD%"
if "%_JAVACMD"=="" (
if not "%JAVA_HOME%"=="" (
if exist "%JAVA_HOME%\bin\java.exe" set "_JAVACMD=%JAVA_HOME%\bin\java.exe"
)
)
if "%_JAVACMD%"=="" set _JAVACMD=java
rem Parses x out of 1.x; for example 8 out of java version 1.8.0_xx
rem Otherwise, parses the major version; 9 out of java version 9-ea
set JAVA_VERSION=0
for /f "tokens=3" %%g in ('%_JAVACMD% -Xms32M -Xmx32M -version 2^>^&1 ^| findstr /i "version"') do (
set JAVA_VERSION=%%g
)
set JAVA_VERSION=%JAVA_VERSION:"=%
for /f "delims=.-_ tokens=1-2" %%v in ("%JAVA_VERSION%") do (
if /I "%%v" EQU "1" (
set JAVA_VERSION=%%w
) else (
set JAVA_VERSION=%%v
)
)
#echo %JAVA_VERSION%
endlocal
echoes 8 or 17 etc.


This will extract the minor part of the version number:
java -version 2>&1 | awk '/version/ {print $3}' | awk -F . '{print $2}'
However, it may be better to extract the major.minor and match on that in case Oracle ever change the version number scheme again e.g.:
java -version 2>&1 | awk '/version/ {print $3}' | egrep -o '[0-9]+\.[0-9]+'


for /f tokens^=2-5^ delims^=.-_+^" %j in ('java -fullversion 2^>^&1') do #set "jver=%j%k%l%m"
This will store the java version into jver variable and as integer
And you can use it for comparisons .E.G
if %jver% LSS 16000 echo not supported version
.You can use more major version by removing %k and %l and %m.This command prompt version.
For .bat use this:
#echo off
PATH %PATH%;%JAVA_HOME%\bin\
for /f tokens^=2-5^ delims^=.-_+^" %%j in ('java -fullversion 2^>^&1') do set "jver=%%j%%k%%l%%m"
According to my tests this is the fastest way to get the java version from bat (as it uses only internal commands and not external ones as FIND,FINDSTR and does not use GOTO which also can slow the script). Some JDK vendors does not support -fullversion switch or their implementation is not the same as this one provided by Oracle (better avoid them).

Related

How to access Java property in terminal

I know I can get all the Java System properties from the terminal using
java -XshowSettings:properties -version
How do I access just one specific java system property?
For example, like "user.name"?
I want to do this in the terminal, not with Java.

Solution as a one liner script. Just change the val variable to the key you want to print:
val='java.library.path'; java -XshowSettings:properties -version 2>&1 | sed -re 's/^ +[^=]+ =/_&/' | gawk -v key=$val 'BEGIN{ RS="_"; IFS=" = "} { if($1 ~ key){ print $0 }}'
Details
Some property values like java.library.path contain new lines so we need to mark records before filtering and printing them.
sed allows us to do that, then awk can be used to filter and print.
java -XshowSettings:properties -version 2>&1 |\
sed -re 's/^ +[^=]+ =/_&/' |\
gawk -v key=java.library.path 'BEGIN{ RS="_"; IFS=" = "} { if($1 ~ key){ print $0 }}'
Result:
java.library.path = /usr/java/packages/lib/amd64
/usr/lib64
/lib64
/lib
/usr/lib
Pipeline parts explained:
2>&1: properties are printed to stderr so we need to redirect them to stdin.
sed -re 's/^ +[^=]+ =/_&/' : add an underscore in front of interesting lines, those starting with 4 spaces and containing =.
gawk -v key=java.library.path: set keyawk variable to the selected property key.
'BEGIN{ RS="_"; IFS=" = "}: set record separator to '_' and input field separator IFS to =.

If you need the current logged in user in bash just use whoami command. If you want to get the java property from terminal you can use the following command
java -XshowSettings:properties -version 2>&1 | grep user.name
which will print
$java -XshowSettings:properties -version 2>&1 | grep user.name
user.name = user
If you just need the user name only
java -XshowSettings:properties -version 2>&1 | grep user.name | cut -c 16-100
which will print
$java -XshowSettings:properties -version 2>&1 | grep user.name | cut -c 16-100
user

You can't.
What you can do is create a java file to get the information and run with java, here the documentation.
Since you already said that you don't want this, you can grep (filter) the output of
java -XshowSettings:properties -version 2>&1 | grep java.home
java.home = /usr/java/jdk1.8.0_112/jre
If you want to know the system properties of a running jvm, use the jcmd tool
jcmd PID VM.system_properties

how to get latest JRE version in command prompt if windows has more than one JRE version

I am trying get the latest JRE installed in windows using dir command. I Have JRE 1.6 and 1.8 installed on my Windows but I need to get whatever the latest version installed in windows (even in future it may change to 1.8 to 2.0*). Can any one please help on this.

Try this batch for win7
#echo off
#echo off
rem http://technet.microsoft.com/en-us/library/bb490890.aspx
SETLOCAL EnableDelayedExpansion
:: findJDK.bat
rem start:-> Run a command in a separate process, or run a file with its default associated application
rem REGEDIT.EXE: -> Export to a (.REG) file:
rem REGEDIT.EXE [ /L:system | /R:user ] /E exportfile.REG "registry_key"
start /w regedit /e reg1.txt "HKEY_LOCAL_MACHINE\SOFTWARE\Wow6432Node\JavaSoft\Java Development Kit"
rem TYPE: -> Display text file content in console
rem FIND: -> Search files or standard output
rem reserved key words regarding reg-key -> "JavaHome" "MicroVersion" "RuntimeLib"
rem | Reads the output from one command and writes it to the input of another command. Also known as a pipe.
type reg1.txt | find "JavaHome" > reg2.txt
if errorlevel 1 goto ERROR
for /f "tokens=2 delims==" %%x in (reg2.txt) do (
set JavaTemp=%%~x
echo Regedit: JAVA_HOME path : !JavaTemp!
)
if errorlevel 1 goto ERROR
echo.
set JAVA_HOME=%JavaTemp%
set JAVA_HOME=%JAVA_HOME:\\=\%
echo JAVA_HOME was found to be %JAVA_HOME%
start /w regedit /e reg3.txt "HKEY_LOCAL_MACHINE\SOFTWARE\Wow6432Node\JavaSoft\Java Runtime Environment"
type reg3.txt | find "RuntimeLib" > reg4.txt
if errorlevel 1 goto ERROR
for /f "tokens=2 delims==" %%x in (reg4.txt) do (
set JavaTemp=%%~x
echo Regedit: JRE_HOME path : !JavaTemp!
)
if errorlevel 1 goto ERROR
echo.
set JRE_HOME=%JavaTemp%
set JRE_HOME=%JRE_HOME:\\=\%
echo JRE_HOME was found to be %JRE_HOME%
goto END
:ERROR
echo A reg1.txt is: & type reg1.txt
echo B reg2.txt is: & type reg2.txt
echo
:END
echo END
pause

java -version:1.8.0_101 -jar jarname.jar
Use above way to invoke java here you can give version of java which you want to use in case there are multiple in your path .

Set Java Major Version Number As A Variable

This is irritating. I am trying to incorporate a trigger in a script to run if the major version of Java is 1.6,1.7, etc.
In the line I am trying to set the variable in, I pull the Java version and pipe it to for with tokens and delims identified, resulting in the "do" as a set command for variable %jver%. However, echoing %jver% results in "echo is on". Why wont it set the variable? Everything looks legit until %jver% is used.
Yes, I double the percentages for the script. The code here is for use at the command prompt.
Here is the line:
%systemroot%\system32\java.exe -version 2>&1 | for /f "tokens=1-4 delims=. " %a in ('findstr /i "version"') do (set jver=%~c.%d)

#rao thank you for helping me discover an alternate solution to meet my intent.
My workaround solution to this is the following line:
for /f "tokens=1-4 delims=. " %a in ('java -version 2^>^&1 ^| findstr /i "version"') do (SET JVER=%~c.%d)

Looks this is answered in this post by Patrick Cuff
Here is the solution adding from mentioned original post
#echo off
setlocal
set VERSION6="1.6.0_21"
for /f "tokens=3" %%g in ('java -version 2^>^&1 ^| findstr /i "version"') do (
#echo Output: %%g
set JAVAVER=%%g
)
set JAVAVER=%JAVAVER:"=%
#echo Output: %JAVAVER%

How do I find where JDK is installed on my windows machine?

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).

If you are using Linux/Unix/Mac OS X:
Try this:
$ which java
Should output the exact location.
After that, you can set JAVA_HOME environment variable yourself.
In my computer (Mac OS X - Snow Leopard):
$ which java
/usr/bin/java
$ ls -l /usr/bin/java
lrwxr-xr-x 1 root wheel 74 Nov 7 07:59 /usr/bin/java -> /System/Library/Frameworks/JavaVM.framework/Versions/Current/Commands/java
If you are using Windows:
c:\> for %i in (java.exe) do #echo. %~$PATH:i

Windows > Start > cmd >
C:> for %i in (javac.exe) do #echo. %~$PATH:i
If you have a JDK installed, the Path is displayed,
for example: C:\Program Files\Java\jdk1.6.0_30\bin\javac.exe

In Windows at the command prompt
where javac

Command line:
Run where java on Command Prompt.
GUI:
On Windows 10 you can find out the path by going to Control Panel > Programs > Java. In the panel that shows up, you can find the path as demonstrated in the screenshot below. In the Java Control Panel, go to the 'Java' tab and then click the 'View' button under the description 'View and manage Java Runtime versions and settings for Java applications and applets.'
This should work on Windows 7 and possibly other recent versions of Windows.

In windows the default is: C:\Program Files\Java\jdk1.6.0_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 HKLM\SOFTWARE\JavaSoft\Java Runtime Environment and HKLM\SOFTWARE\Wow6432Node\JavaSoft\Java 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.

Plain and simple on Windows platforms:
where java

Under Windows, you can use
C:>dir /b /s java.exe
to print the full path of each and every "java.exe" on your C: drive, regardless of whether they are on your PATH environment variable.

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="HKLM\SOFTWARE\JavaSoft\Java 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="HKLM\SOFTWARE\JavaSoft\Java 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

In a Windows command prompt, just type:
set java_home
Or, if you don't like the command environment, you can check it from:
Start menu > Computer > System Properties > Advanced System Properties. Then open Advanced tab > Environment Variables and in system variable try to find JAVA_HOME.

Powershell one liner:
$p='HKLM:\SOFTWARE\JavaSoft\Java Development Kit'; $v=(gp $p).CurrentVersion; (gp $p/$v).JavaHome

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 Files\Oracle\Java\jav...
Application java.exe 8.0.131... C:\ProgramData\Oracle\Java\javapath\java.exe

Run this program from commandline:
// File: Main.java
public class Main {
public static void main(String[] args) {
System.out.println(System.getProperty("java.home"));
}
}
$ javac Main.java
$ java Main

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.

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

Just execute the set command in your command line. Then you see all the environments variables you have set.
Or if on Unix you can simplify it:
$ set | grep "JAVA_HOME"

This is OS specific. On Unix:
which java
will display the path to the executable. I don't know of a Windows equivalent, but there you typically have the bin folder of the JDK installation in the system PATH:
echo %PATH%

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

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.

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="HKLM\SOFTWARE\JavaSoft\Java Development Kit"
SET KEY_2=HKLM\SOFTWARE\JavaSoft\JDK
SET REG_1=reg.exe
SET REG_2="C:\Windows\sysnative\reg.exe"
SET REG_3="C:\Windows\syswow64\reg.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:\Windows\sysnative"
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:\Windows\syswow64"
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

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

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)

in Windows cmd:
set "JAVA_HOME"

#!/bin/bash
if [[ $(which ${JAVA_HOME}/bin/java) ]]; then
exe="${JAVA_HOME}/bin/java"
elif [[ $(which java) ]]; then
exe="java"
else
echo "Java environment is not detected."
exit 1
fi
${exe} -version
For windows:
#echo off
if "%JAVA_HOME%" == "" goto nojavahome
echo Using JAVA_HOME : %JAVA_HOME%
"%JAVA_HOME%/bin/java.exe" -version
goto exit
:nojavahome
echo The JAVA_HOME environment variable is not defined correctly
echo This environment variable is needed to run this program.
goto exit
:exit
This link might help to explain how to find java executable from bash: http://srcode.org/2014/05/07/detect-java-executable/

Script for 32/64 bit Windows.
#echo off
setlocal enabledelayedexpansion
::- Get the Java Version
set KEY="HKLM\SOFTWARE\JavaSoft\Java Runtime Environment"
set KEY64="HKLM\SOFTWARE\WOW6432Node\JavaSoft\Java 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="HKLM\SOFTWARE\JavaSoft\Java Runtime Environment\%JRE_VERSION%"
set KEY64="HKLM\SOFTWARE\WOW6432Node\JavaSoft\Java 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

How do I detect 64-bit Java from the command line?

Is there any way to get at Java internal properties, such as sun.arch.data.model , from a command line on windows? I need a command to put in a batch script that will detect the java architecture type: 32-bit or 64-bit .

If you are using Sun's VM (and I would suppose other VMs have similar details in their version information), you can check for the string "64-Bit" in the output of "java -version":
java -version 2>&1 | find "64-Bit" >nul:
if errorlevel 1 (
echo 32-Bit
) else (
echo 64-Bit
)

jarnbjo's script is for Windows. In Unix shell, you can use the following script.
#!/bin/sh
BIT=`java -version 2>&1`
case "$BIT" in
*64-Bit*)
echo "64-Bit"
;;
*)
echo "32-Bit"
;;
esac

Here is a prewritten property dump program for you: linky

If you install Groovy you can use
groovy -e "System.properties.each{println it}"
for all properties, and
groovy -e "println System.properties['sun.arch.data.model']"
for specific properties.
Installing Groovy is as easy as extracting a zip and add to path.

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