CODEWITHSUNDEEP
windows
Simran Kaur
Simran Kaur

Reputation: 1101

C:/Program is not recognized as an internal or external command

I have JAVA_HOME variable in System variables and is set to

C:\Program Files\Java\jdk1.8.0_60

I get an error when I try to use command %JAVA_HOME%

C:/Program is not recognized as an internal or external command.

I understand it is because of space in the path after Program,

I have also tried using quotes across the path but it did not resolve the issue.

i.e

"C:\Program Files\Java\jdk1.8.0_60"

but no help.

Edit: I tried removing ones with Program Files at all.

Now my System path variable looks like this:

%SystemRoot%\system32;%SystemRoot%;%SystemRoot%\System32\Wbem;%SYSTEMROOT%\System32\WindowsPowerShell\v1.0\;

Environment path variable looks like this:

 C:\Users\dev4\AppData\Roaming\npm; C:\mongodb\bin;

So, I do not really see any of these with space or program files in path and yet when I try

%Path%

it gives the same error.

EDIT 2:

cd %PATH% gives The file name or extension is too long.

however cd %JAVA_HOME% works

Edit 3: Now, the PATH variable has location fo JDK bin but none of my java commands are recognized

Upvotes: 1

Views: 12576

Answers (2)

Sunil Kumar
Sunil Kumar

Reputation: 849

For all general case use the short folder path.

the short form folder name can be obtained by executing cmd prompt and visiting the parent directory and type dir /x and press enter you will see the short form of sub folder names

ex:- C:/Progra~1/MARIAD~1.2/bin/mysql.exe in the place of 

C:/Program Files/MariaDB 10.2/bin/mysql.exe

here short form of MariaDB 10.2 can be obtained form visiting c:/program files/ in cmd prompt and type dir /x

Upvotes: 0

Karl
Karl

Reputation: 3372

wrapping the variables in double quotes should work

"%JAVA_HOME%\bin\java"

Or if you host supports it use a short name

C:\Progra~1\Java\jdk1.8.0_60

Upvotes: 4

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