On Windows, why java -version return Error: opening registry key 'Software\JavaSoft\Java Runtime Environment'?

Question

1. I've removed all ancient Java 8 install on my computer.
2. I've downloaded latest openjdk 11 from https://jdk.java.net/11/ and extracted the content to C:\Program Files\Java\.

3. I've setup system environment variable to

   JAVA_HOME = C:\Program Files\Java\jdk-11.0.1
   

   and added %JAVA_HOME%\bin to Path.

4. In a command prompt > java -version returns:

   Error: opening registry key 'Software\JavaSoft\Java Runtime Environment'
   Error: could not find java.dll
   Error: Could not find Java SE Runtime Environment.
   

5. Why?

Answer 1

How to make it work:

Windows > Start > cmd >

C:> for %i in (javac.exe) do @echo.   %~$PATH:i

or

C:> where java

It returns C:\ProgramData\Oracle\Java\javapath folder which contain:

• java.exe
• javaw.exe
• javaws.exe

Browse to this folder with Windows file explorer and remove everything (three files above).

Close and reopen cmd (Windows > Start > cmd >)

C:> java -version

Should now return:

openjdk version "11.0.1" 2018-10-16

OpenJDK Runtime Environment 18.9 (build 11.0.1+13)

OpenJDK 64-Bit Server VM 18.9 (build 11.0.1+13, mixed mode)

or something similar depending on which java release package you've downloaded...

It works!

Answer 2

I came across same issue,

how to check:

1. open cmd, execute "where java" (without quote)
2. you should see more than one output of the java path, one of it should be the one you set "C:\Program Files\Java\jdk-11.0.1\bin", the other ones should be in above

how to solve:

option 1:

simply delete all the other java path

option 2:

check your enviroment variable setting, move your %JAVA_HOME%\bin to the above all the other java path

Answer 3

Try Adding the new Java home path at the start of the path instead of end i.e %JAVA_HOME%\bin;(rest of path). It will get picked first. Worked for me.