How to open a file by clicking a button in Python 3

Question

I want to open a file (HTML web document) by clicking a button on my menu bar. I am using Python 3.4 and Windows 7 64-bit. How would I go about doing this?

The HTML document is saved on my computer, I want it open it from my computer.

Answer 1

For creating button in Python use Tkinter widget.

import Tkinter
import tkMessageBox

top = Tkinter.Tk()
def helloCallBack():
tkMessageBox.showinfo( "Hello Python", "Hello World")
B = Tkinter.Button(top, text ="Hello", command = helloCallBack)
B.pack()
top.mainloop()

Answer 2

I do not know how event handlers for a button work in Tkinter, but the python command to open an html link or file is:

import webbrowser
webbrowser.open(file_path)

This opens the link or file in the default browser. Here's the documentation .

Answer 3

Well, You can set an event handler attached to that button on your menu bar calling a function like this ..

from tkinter import filedialog as fd, messagebox as mb
from webbrowser import open as openlink

def openHTML(file = None):
    if file is None: # browse file in filedialog
        file = fd.askopenfilename(filetypes=(("All Files", "*.*"), ("HTML Documents", "*.html;*.htm"))
    if not file: # if the user didn't cancel
        return
    try: # trying to access the file, if it isn't encrypted or something ..
        open(file, 'r')
    except EnvironmentError as e:
        mb.showerror(title = 'Something isn\'t good ..', detail = e.message)
        return
    openlink(file) # finally, opening the document