How to open a directory path with spaces in Python?

Question

I am trying to pass source and destination path from the command line as below

python ReleaseTool.py -i C:\Users\Abdur\Documents\NetBeansProjects\Exam System -o C:\Users\Abdur\Documents\NetBeansProjects\Release

but it is throwing error

WindowsError: [Error 3] The system cannot find the path specified: ''

due to 'Exam System' which has a space between. Please suggest how to handle this.

try put `""` in your path. For example: `python ReleaseTool.py -i "C:\Users\Abdur\Documents\NetBeansProjects\Exam System" -o C:\Users\Abdur\Documents\NetBeansProjects\Release ` — Avihoo Mamka, Nov 15 '15 at 07:54
Try this: `python ReleaseTool.py -i C:\Users\Abdur\Documents\NetBeansProjects\"Exam System" -o C:\Users\Abdur\Documents\NetBeansProjects\Release` — narendra-choudhary, Nov 15 '15 at 08:00

score 6 · Answer 1 · answered Nov 15 '15 at 07:57

Cause

Long filenames or paths with spaces are supported by NTFS in Windows NT. However, these filenames or directory names require quotation marks around them when they are specified in a command prompt operation. Failure to use the quotation marks results in the error message.

Solution

Use quotation marks when specifying long filenames or paths with spaces. For example, typing the following at the command prompt copy c:\my file name d:\my new file name results in the following error message:

The system cannot find the file specified.

The correct syntax is: copy "c:\my file name" "d:\my new file name" Note that the quotation marks must be used.

How to open a directory path with spaces in Python?

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