'keytool' is not recognized as an internal or external command

Question

I've created a small Android application using Ionic and I was going to build the signed apk to test the app.

When I typed 'keytool' command in my command prompt I'm getting this error

'keytool' is not recognized as an internal or external command, operable program or batch file.

And I tried googling and all solved problems mentions that it's a problem with Path variable of JDK. I tried changing my 'JAVA_HOME' variable value to 'C:\Program Files\Java\jdk1.8.0_131\bin' and to 'C:\Program Files\Java\jre1.8.0_131\bin' and still getting the same error.

But when I run the 'keytool' command it the cmd 'C:\Program Files\Java\jre1.8.0_131\bin' location, it works.

Can anyone help me with this ? What might be a problem that I can't access it globally ?

Solution

That happens because you haven't added the bin directory of your JAVA_HOME environment variable in your %PATH%.

1. Add the environment variable JAVA_HOME with path C:\Program Files\Java\jdk1.8.0_131 without the bin directory.
2. Add the %JAVA_HOME%\bin directory at the end of your %PATH%.

Comments

[Stephen C]

The problem is that the bin directory is not on your %PATH%. The command shell (e.g. "cmd.exe") looks for commands in the search path given by %PATH%.

[Thidasa Pankaja]

Thanks. But how to add the bin directory to %PATH% ?

[Thidasa Pankaja]

Thanks. As you've mentioned I changed JAVA_HOME variable path to 'C:\Program Files\Java\jdk1.8.0_131' But I don't get the 2nd point

[Giorgio Antonioli]

@ThidasaParanavitharana You have also a variable named Path. You can find inside it, some paths to executable. At the end of these paths, you have to add the last path with %JAVA_HOME%\bin

[Thidasa Pankaja]

Thanks a lot. It worked

[Giorgio Antonioli]

@ThidasaParanavitharana No problem. If it worked, mark the answer as accepted

[Matias]

Perfect explanation. Just what I needed. Thanks a lot!