How to fix '.' is not an internal or external command error

user_1620040 0
Viewed 140750

I have followed few links to try and solve this issue, such as link1 , where they have asked to me include the path of the exe in the environment variables. This is the following command I tried, to get this error

       D:\Gesture Recognition\Gesture Recognition\Debug>./"Gesture Recognition.exe" 
       rawrec1.trr

and the error

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

I included the gesture recognition exe path into the systems and user variables of the environment variables. Even after booting the system, the error still persisted. Can anybody help me to solve this or fix this? Thanks in advance

cmd environment-variables
6 Answers

This error comes when using the following command in Windows. You can simply run the following command by removing the dot '.' and the slash '/'.

Instead of writing:

D:\Gesture Recognition\Gesture Recognition\Debug>./"Gesture Recognition.exe"

Write:

D:\Gesture Recognition\Gesture Recognition\Debug>"Gesture Recognition.exe"

user_131224630

I also encountered this issue when working with Webpack. If packages are installed with NPM, ideally you should not have to provide a path to the module that you intend to run. Something like this should work:

"build": "webpack --config webpack.config.js",
"start": "webpack serve --config webpack.config.js --open"

From my testing, this works in Git Bash, Powershell, and the Command Prompt.

user_166458110

I cloned a repo from github on my windows machine and there command to start cypress was

"cypress:open": "../../node_modules/.bin/cypress open"

and this was giving me this error

. . is not recognized as internal or external command

I don't know may be on linux, command may require node_modules path as well but on my windows machine just

"cypress:open": "cypress open"

was sufficient

user_55686460
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