I have recently uninstalled JDK 11 and installed JDK 8. For confirmation, I want to check which JDK is installed on my Windows 10 machine. I typed java -version on cmd then get the error message
java is not recognized as an internal or external command
How to know which JDK version installed on my PC?
you might need to add path in environment variables which you can find in Control Panel
open the Jdk where you installed and add until /bin in the path in environment variables.
Add until /bin in path variable in System Variables which is residing in Environment Variables.
Then do
java -version
which might show up.
If still problem persists, try restarting your pc and see.
You need to update your Windows path to include your %JAVA_HOME%\bin directory. %JAVA_HOME% is the directory that you installed Java into and is also an environment variable that you need to configure for command line execution of your applications. You can edit both of these in the Windows control panel and you should restart.
When you run java -version you will see the internal version number. This is explained here: https://en.wikipedia.org/wiki/Java_version_history.
Basically, you can ignore the 1. when reading version number. The _xxx is a reference to the most recent patch or build release.
On Windows 10, this required mapping the environment variable for JAVA_HOME to the JDK installation directory. Use these steps:
Run the installer for the JDK. (available for windows here: https://www.oracle.com/java/technologies/downloads/#jdk17-windows)
windows key -> Environment Variables, select the only result
In the System Properties window that opened, select Environment Variables
Select new button under the User variables section
Variable name: JAVA_HOME, Variable Value: <The JDK filepath from step 0>
ok all open menus
Close any open cmd prompt windows
open a new cmd window and type echo %JAVA_HOME% It should print the installation path for the JDK.
To get your jdk location in Windows, run this at a command prompt:
where java
This lists any and all locations of java.exe, including from your JAVA_HOME. For example, the 3rd line here reflects my JAVA_HOME location, where I'm pointing to JDK 8:
C:\Users\me> where java
C:\Program Files\Common Files\Oracle\Java\javapath\java.exe
C:\Program Files (x86)\Common Files\Oracle\Java\javapath\java.exe
C:\Program Files\Java\jdk1.8.0_202\bin\java.exe
Note for comparison that java -version does not reflect my JAVA_HOME location and in fact shows java version 11 instead of 8:
C:\Users\me> java -version
java version "11.0.15" 2022-04-19 LTS
Java(TM) SE Runtime Environment 18.9 (build 11.0.15+8-LTS-149)
Java HotSpot(TM) 64-Bit Server VM 18.9 (build 11.0.15+8-LTS-149, mixed mode)
This is confusing because my Java compiles (e.g., via mvn) use JDK 8 since that's what my JAVA_HOME is pointing to. (I'm not even sure where the version 11 it found came from; possibly from when I installed maven.)
Determining the difference between the JRE and JDK you're running has never been straightforward. Seems like java -version used to be a way to do this, but no longer.
Adding to the complexity, you can also supposedly get your Java version info from Control Panel > Programs > Java > About. For me, that shows Version 8. That's despite java -version showing version 11.0.15. And it doesn't change even if I point my JAVA_HOME to JDK 11.
Note that this answer is also helpful. In my case, that helped me determine that I have java.exe and javac.exe at C:\Program Files (x86)\Common Files\Oracle\Java\javapath and C:\Program Files\Common Files\Oracle\Java\javapath. Depending on which one I have listed first in my Path variable, I get different results when i run java -version or java --version. The former seems to work when Java 8 is listed first; the latter when Java 11 is first.
Related
I have a Windows 10 x64 machine. I had Open JDK 10. My JAVA_HOME points to Open JDK 10 location. My Path variable has %JAVA_HOME%/bin; part.
Then I installed Java 8.
https://www.oracle.com/technetwork/java/javase/downloads/jdk8-downloads-2133151.html
Java installer changes Path.
C:\Users\user>set Path
Path=C:\Program Files (x86)\Common Files\Oracle\Java\javapath;OTHER_STUFF_WITHOUT_JAVA
Now in command line java -version points to Java 8.
C:\Users\user>java -version
java version "1.8.0_202"
Java(TM) SE Runtime Environment (build 1.8.0_202-b08)
Java HotSpot(TM) 64-Bit Server VM (build 25.202-b08, mixed mode)
But JAVA_HOME is still pointing to Open JDK 10.
C:\Users\user>set java_home
JAVA_HOME=C:\Java\jdk\jdk-10.0.2
My goal is to have Open JDK-10 as default Java.
This is not a duplicate
Default Java Path on Windows Machines
because there is no answer for me.
(Post comment as the answer for readability)
The easiest way to have several JDK and be sure to point towards the good one, is to avoid using the java installers. You should just unzip the JDKs in a folder and then modify your JAVA_HOME and PATH accordingly. You could also have several JAVA_HOME like JAVA_HOME_8 and JAVA_HOME_10 pointing to their respective JDK folder and make your JAVA_HOME=%JAVA_HOME_10% so you can change easily.
The easiest way i just found out is that the java 8 installer creates a new environment PATH entry:
C:\Program Files (x86)\Common Files\Oracle\Java\javapath
This will point to the latest installed JRE or JDK(i.e. the last in JDK or JRE) using the installer method.
So if your JAVA_HOME points to the JDK 10 OR JRE 10 just go to the Environment Variables and remove:
C:\Program Files (x86)\Common Files\Oracle\Java\javapath
From the PATH variable.
Then the values you used for JAVA_HOME will become effective.
This worked for me after installing java 8 and i had installed java 12
I am using the 'jshell command in my machine it is not recognised. But java command is working fine. is there any environment setup for jshell in jdk 10
C:\Users\Kannan
λ jshell
'jshell' is not recognized as an internal or external command,
operable program or batch file.
C:\Users\Kannan
λ java -version
java version "10" 2018-03-20
Java(TM) SE Runtime Environment 18.3 (build 10+46)
Java HotSpot(TM) 64-Bit Server VM 18.3 (build 10+46, mixed mode)
jshell is a part of JDK 10 and it's located in the %JAVA_HOME%\bin folder on Windows.
Possible problems:
You installed JRE 10 only (instead of JDK 10). jshell is NOT part of the JRE.
%JAVA_HOME%\bin is not part of the PATH system variable.
See also: Environment variables for java installation
check if jshell is installed with your Java environment. ls [JAVA-INSTALLEDPATH]/bin
if jshell does not exist download appropriate JDK
if jshell is present append path in your environment profile or use full path.
You need to add the bin folder of your Java Development Kit (JDK) installation to the PATH environment variable. The java command works, because the JRE installs a copy of the java.exe executable in C:\ProgramData\Oracle\Java\javapath\ and adds it to the PATH.
For editing the PATH, see How do I set system environment variables in Windows 10? on superuser
Had the same problem. I set JAVA_HOME and wrapped it with double quotes ("").
JAVA_HOME="C:\Program Files\Java\jdk1.8.0_144"
What you have is a JRE. You should install JDK and set JAVA_HOME. Then jshell will start working properly.
The same problem has happened to me.
Uninstall and Install the right JDK from Oracle website.
steps:
1. Control Panel > System & Security > System > Change Setting > Advance > Change Environment Variable > System Variable > New > Variable Name:"Path" & Variable Value: "C:\Program Files\Java\jdk-10.0.1\bin" (address of the bin)> ok
You are good to go.
Here's the link to the video that helped me.
https://www.youtube.com/watch?v=UokTaTwckDw
Run you command prompt as Administrator. window -> cmd -> Run as Administrator then type jshell.
Remove any default path variable that is set in envirnoment variables when you install JDK apart from JAVA_HOME.Then update your JAVA_HOME to /path of JDK 9 OR Above/
I normally still use Java 7 for all my coding projects (it's a company "politics" issue), but I installed Java 8 for one third-party project I am contributing to. Now, it seems I cannot have Java 8 installed in Windows 7 x64, and still use Java 7 by default:
C:\>"%JAVA_HOME%\bin\java.exe" -version
java version "1.7.0_55"
Java(TM) SE Runtime Environment (build 1.7.0_55-b13)
Java HotSpot(TM) 64-Bit Server VM (build 24.55-b03, mixed mode)
C:\>java.exe -version
java version "1.8.0_05"
Java(TM) SE Runtime Environment (build 1.8.0_05-b13)
Java HotSpot(TM) 64-Bit Server VM (build 25.5-b02, mixed mode)
As you can see, JAVA_HOME is completely ignored.
I also have Java in the path, using "%JAVA_HOME%\bin", which resolve correctly to Java 7 when I check the path in a DOS box, but it still makes no difference.
I checked in the "Java Control Panel" (not sure if this affects the default command-line Java version). Under the "Java" tab, the "View..." button, you get to see "registered" Java versions. I can add all the versions under the "User" tab, but under "System" there is only Java 8, and no way to change it.
Am I missing something, or did Oracle just make it impossible to use Java 7, unless I de-install Java 8? I don't want to have to specify the "source" and "target" everywhere, and I don't even know if it is possible for me to specify it everywhere, where Java is used.
EDIT: What I did is I de-installed all Java. Then installed the latest Java7 (both 86 and x64), and then the latest Java8 (both 86 and x64). After I did that, I noticed that the x64 JDK was gone. It seems Java8 killed it. So I re-installed the JDK 7 x64, after the JDK 8 x64. Still, JDK7 x64 did not seem to "replace" the "java.exe" which is copied into the "Windows" directory itself (I assume THAT is the problem).
When you install jdk8 it adds an entry like this
C:\ProgramData\Oracle\Java\javapath
to beginning of your PATH environment variable, removing this entry should resolve your problem.
You can select the JRE version from the command line with the -version: option.
> java -version:"1.7" MyClass
should select the 1.7 JRE if installed properly.
The list of the properly installed JRE is in the registry, see the key :
HKEY_LOCAL_MACHINE\SOFTWARE\JavaSoft\Java Runtime Environment (32bit)
HKEY_LOCAL_MACHINE\SOFTWARE\Wow6432Node\JavaSoft\Java Runtime Environment (64bit)
You can set the "CurrentVersion" there if you want a different default version than the latest.
See http://docs.oracle.com/javase/7/docs/technotes/tools/windows/java.html#options
Don't modify your PATH to point to a particuliar JRE, let the special java.exe in Windows/system32 do the job.
Windows and Unix both find programs using their PATH environment variable. You have an java.exe in your Windows\System32 which is appearing before your "preferred" version of Java.
Change the PATH to be the one you need, or specify the full path when you need a different version.
2 Steps
1
Change registry key **HKEY_LOCAL_MACHINE\SOFTWARE\JavaSoft\Java Runtime Environment\CurrentVersion** to point to 1.7
2
Copy java.exe,javaw.exe and javaws.exe from your java 1.7 version to Windows\System32 folder
(Since the corresponding files of java 1.8 are already there, you might have to overwrite with admin permissions)
3
(OOps actually not required 3rd step )
Open a new cmd window and check
java -version
Looks like you have to check where in your PATH is located your JAVA_HOME variable, the PATH is evaluated from left to right. A tip for you is to do all your Java system variables configuration at the beginning of your PATH.
PATH = %M2_HOME%\bin;%JAVA_HOME%\bin;C:\ProgramData\Oracle\Java\javapath;...
Probably that's why after doing this:
- java -version
you are getting this:
- java version "1.8.0_05"
because there are other areas in your PATH that are pointing to other java.exe, for example C:\Windows\System32 or C:\ProgramData\Oracle\Java\javapath etc.
I had to make 2 changes for it to work:
Changed the Registry key 'Software\JavaSoft\Java Runtime Environment'\CurrentVersion' to 1.7 from 1.8
The Java 8 installation adds a new entry to the PATH environment variable 'C:\ProgramData\Oracle\Java\javapath'. I removed this entry from the PATH.
I have two javas installed on my pc. jdk1.7.0_45 (x64) and jdk1.6.0_45(x86). I want that default java on my machine would be jdk1.6.0_45(x86). I added JAVA_HOME with value C:\Program Files (x86)\Java\jdk1.6.0_45 and added C:\Program Files (x86)\Java\jdk1.6.0_45\bin to system path, but still when i type in command prompt "java -version" it says:
java version "1.7.0_45"
Java(TM) SE Runtime Environment (build 1.7.0_45-b18)
Java HotSpot(TM) 64-Bit Server VM (build 24.45-b08, mixed mode)
I have no idea why the things are like that, because my OS should see jdk1.7.0_45 (x64) version. Why is that so?
I use windows 8
edit:
after setting java home to jdk1.7.0_45 (x64) folder and updating path accordingly javac works, but then I set java home and path to jdk1.6.0_45(x86) javac does not work.
You need to add the %JAVA_HOME%\bin to the very beginning of your path. This is because Java is installed in your %SystemRoot%\system32, too. If you add the %JAVA_HOME%\bin to the beginning of your path the Java installed in your JAVA_HOME will be used.
JAVA_HOME does not help when we run java from command line, you should fix PATH system var
I added JAVA_HOME with value C:\Program Files (x86)\Java\jdk1.6.0_45
In Windows, you should excape spaces in the path: JAVA_HOME="C:\Program Files (x86)\Java\jdk1.6.0_45". But the best way is to set Java to a folder that doesn't have spaces in its path. For example, I use C:\Java\Java6
and added C:\Program Files (x86)\Java\jdk1.6.0_45\bin to system path
Probably you added after path to the Java 7. Make sure that Java 7 doesn't exist in the system path.
I've installed jdk 7 and set up the "JAVA_HOME" to this version. Then, I've installed jdk 6 and I've updated the "JAVA_HOME" to point to jdk 6. After closing and reopening the system, when typing "java -version", I got always :
java version "1.7.0_40"
Java(TM) SE Runtime Environment (build 1.7.0_40-b43)
Java HotSpot(TM) Client VM (build 24.0-b56, mixed mode, sharing)
However, when typing
echo %JAVA_HOME%
I got :
C:\Program Files\Java\jdk1.6.0_45
I didn't understand why java -version don't display the value of "JAVA_HOME"
Java 7 puts a java.exe in c:\windows\system32. You could try to delete these exe's, but I'm not sure Windows will allow it, or restore it after a reboot.
If you want to override it, you must put %JAVA_HOME%\bin as first entry in your PATH, before c:\windows\system32.
Which java starts when you run java -version depends on the PATH env variable, not JAVA_HOME. OS will be looking for java.exe (Windows) like for any other program
if you run java on your command window, it takes java from the PATH variable. JAVA_HOME is used my maven etc...
I believe your PATH variable is pointing to JAVA version 1.7.0_40/bin directory.
According to my understanding, this is nothing to do with your JAVA_HOME environment variable. You are getting java version from C:\windows\system32\java.exe.
If you don’t want that behaviour then in system variables section put %JAVA_HOME%\bin as starting element (Of course you should have JAVA_HOME). (I thought of adding image but I dont have enough reputation to do so :( )