Reputation: 661
ok so I got a new PC
Windows 7 Pro 64 bit Service Pack 1
I installed jdk-7u25-windows-x64 ( jdk 1.7.0_25)
It is installed in default location C:\Program Files\Java\jdk1.7.0_25
I set the JAVA_HOME environment variable to C:\Program Files\Java\jdk1.7.0_25
I added %JAVA_HOME%\bin to the Path environment variable (yes I used a ; before I added it)
Clicked Ok
Closed all CMD windows opened them up
And now If I run the command java I get the expected output
But If I run the javac command I get this
C:\Users\AJ>javac
'javac' is not recognized as an internal or external command,
operable program or batch file.
I have tried Fiddling with the path but no luck, and have restarted my PC etc. no luck
Also it seems to work if I am in the jdk\bin directory
C:\>cd C:\Program Files\Java\jdk1.7.0_25\bin
C:\Program Files\Java\jdk1.7.0_25\bin>javac -version
javac 1.7.0_25
C:\Program Files\Java\jdk1.7.0_25\bin>
how do I get javac command to run correctly like java is doing
Upvotes: 2
Views: 12278
Reputation: 41
You can also try to specify the full pathname.
Instead of
%JAVA_HOME%\bin
try
C:\Program Files\Java\jdk1.8.0_101\bin
This worked for me on Windows 10 64-bit Education Edition with Java 8.0.101
Upvotes: 4
Reputation: 438
I had previously problems with the path containing a blank. I would suggest to change the path from
C:\Program Files\Java\jdk1.7.0_25
to
C:\Progra~1\Java\jdk1.7.0_25
Hope this solves your problem!
Upvotes: 7