Python/Django :How to open file in xlrd using file object.

Question

Background:

I have uploaded a xls file using the FileField. Now I want to parse this file using the xlrd tool.

While doing so I am doing the following in the view.

if form.is_valid():
            user_file = form.save()
            user_file.save()             
            workbook = xlrd.open_workbook(user_file.file.name)//( user_file.file.name returns relative path to the media_root folder. )
            sheet = workbook.sheet_by_index(0)

This gives an error as xlrd is unable to find the file.(As the path is not absolute). How can I do this ?

score 2 · Accepted Answer · answered Dec 21 '12 at 10:17

2

os.path.join(settings.MEDIA_ROOT, user_file.file.name)

Should give the correct path.

answered Dec 21 '12 at 10:17

Thomas Orozco

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Oh... Right my bad ... Thanks for the quick response. – Akash Deshpande Dec 21 '12 at 10:20

Python/Django :How to open file in xlrd using file object.

1 Answers1