CODEWITHSUNDEEP
batch-filecmd
DataPsycho
DataPsycho

Reputation: 988

timeout in Batch after changing path is not working?

I am new to cmd. I have created a file to create folders and sub folders. I have put some timeout. The first timeout is working but other time out is not working do not know why. Is it because of changing the path. I have checked line by line in CMD. In CMD it works fine.

Input:

timeout /t 5

@echo off
set path=C:\Users\bruce\Documents\Project-Gallary
cd %path%
@echo "Welcome master Bruce !"
@echo "Recommended Project Folder Name: G01-Global-Report/C01-Compliance"
echo %PATH%
timeout /t 10
@echo " *** Try not the put space in Folder Name"
set /p ProjectName="What is the project name? "

Output:

C:\Users\bruce\Desktop>timeout /t 5

Waiting for 0 seconds, press a key to continue ...
"Recommended Project Folder Name: G01-Global-Report/C01-Compliance"
C:\Users\bruce\Documents\Project-Gallary
'timeout' is not recognized as an internal or external command,
operable program or batch file.
" *** Try not the put space in Folder Name"
What is the project name?

Upvotes: 1

Views: 2271

Answers (2)

user7818749
user7818749

Reputation:

Big mistake, you are changing a system variable PATH

System variable path lists all the paths to executables, you changed it by setting set path=C:\Users\bruce\Documents\Project-Gallary and now your other cmd's are no longer found.

Try this version instead, first close all your cmd.exe windows to clear all set commands.

@echo off
set "mypath=C:\Users\bruce\Documents\Project-Gallary"
cd %mypath%
@echo "Welcome master Bruce !"
@echo "Recommended Project Folder Name: G01-Global-Report/C01-Compliance"
echo %PATH%
timeout /t 10
@echo " *** Try not the put space in Folder Name"
set /p ProjectName="What is the project name? "

Edit

Based on @npocmaka's comment. If you actually want to add the path to the system path variable then simply replace the set path line to:

set "PATH=%PATH%;C:\Users\bruce\Documents\Project-Gallary"

Upvotes: 6

Compo
Compo

Reputation: 38623

The answer to your question is yes, it is because you have changed, path.

The environment variable %Path% contains locations which are searched when a 'pathless' executable file is not located in the current working directory.

The system also holds another variable called %PATHEXT% which contains a list of executable file extensions which are suffixed to a file name, (in that lists order), if a file without an extension is called and not found.

If you want to see the values of those two variables, open a command prompt and enter, Set Path.

When your script gets to the second Timeout command, it searches the current working directory for any file named Timeout and with any of the extensions listed in %PATHEXT%. If not found, it would similarly search each of the locations in %Path%, which in your case would be C:\Users\bruce\Documents\Project-Gallary; (as timeout was not found you received the error message)

The Timeout command is really timeout.exe which should be located in \Windows\System32. Had you not changed the %Path% variable that location, (preceded by the system drive letter), was defined within the value of the %Path% environment variable and the appropriate executable file found.

Your solutions are therefore to avoid setting variables named %Path% or %PATHEXT% unless absolutely necessary, e.g. Set "lPath=C:\Users\bruce\Documents\Project-Gallary". You could alternatively ignore that advice and use absolute paths with extensions, e.g. C:\Windows\System32\timeout.exe /t 10. Finally you could, if the script requires it, integrate a location, (or extension), into those exiting semicolon delimited variable listings, e.g. Set "Path=%Path%;C:\Users\bruce\Documents\Project-Gallary".

Upvotes: 1

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