How to open a file for browsing (as if one opened it manually)

Question

I am trying to use Python to open a file for visual browsing, i.e. as if one double clicked a file to view it. I have tried numerous searches but because the words are very similar with doing file I/O I could not come across the appropriate information.

I think this is a very simple question / answer and I apologize if the answer was right in front of my nose.

Ideally it would just be a call on a given file path and Python would know the appropriate application to pair in case it was an extension like .pdf.

You mean use Python to invoke Adobe Reader to open a pdf file, for instance? — Stephen Lin, Aug 01 '14 at 09:29
This Question is a duplicate. Check: [http://stackoverflow.com/questions/434597/open-document-with-default-application-in-python][1] [1]: http://stackoverflow.com/questions/434597/open-document-with-default-application-in-python — Christoph Hegemann, Aug 01 '14 at 09:33
http://stackoverflow.com/questions/3277503/python-read-file-line-by-line-into-array — Mher, Aug 01 '14 at 09:34
Many apologies for duplicating this question. I think I lacked the proper vocabulary for describing this--all of what I pulled up before was File I/O. — user3898238, Aug 02 '14 at 01:00

score 0 · Answer 1 · answered Aug 01 '14 at 09:55

0

On Windows you can use os.startfile().

os.startfile("C:\\Users\\your\\file.pdf")

answered Aug 01 '14 at 09:55

TheDarkTurtle

413
1
3
12

Constraint is ONLY WINDOWS :( – Yogeesh Seralathan Aug 01 '14 at 15:58

score 0 · Accepted Answer · answered Aug 01 '14 at 15:57

os.startfile()
os.system() but the parameters passed varies according to the OS

Windows: Start fileName Mac: open fileName Linux: oowriter fileName : gnome-open fileName : kde-open fileName etc...

Example:

fileName="file.pdf" //Path to the file
1. os.startfile(fileName)
2. os.system("start %s")%fileName

How to open a file for browsing (as if one opened it manually)

2 Answers2