Running a Tkinter form in a separate thread

Question

I have written a short module that can be passed an image and simply creates a Tkinter window and displays it. The problem that I am having is that even when I instantiate and call the method that displays the image in a separate thread, the main program will not continue until the Tkinter window is closed.

Here is my module:

import Image, ImageTk
import Tkinter


class Viewer(Tkinter.Tk):
    def __init__(self,parent):
        Tkinter.Tk.__init__(self,parent)
        self.parent = parent
        self.initialize()

    def initialize(self):
        self.grid()

    def show(self,img):
        self.to_display = ImageTk.PhotoImage(img)
        self.label_image = Tkinter.Label(self,image=self.to_display)
        self.label_image.grid(column = 0, row = 0, sticky = "NSEW")
        self.mainloop()

It seems to work fine, except when I call it from my test program like the one below, it will not seem to allow my test program to continue, even when started in a different thread.

import Image
from viewer import Viewer
import threading

def showimage(im):
    view = Viewer(None)
    view.show(im)

if __name__ == "__main__":
    im = Image.open("gaben.jpg")
    t = threading.Thread(showimage(im))
    t.start()
    print "Program keeps going..."

I think that perhaps my problem is that I should be creating a new thread within the module itself, but I was wanting to just try and keep it simple, as I am new to Python.

Anyway, thanks in advance for any assistance.

edit: To clarity, I am just trying to make a module that will display an image in a Tkinter window, so that I can use this module any time I want to display an image. The problem that I am having is that any time a program uses this module, it cannot resume until the Tkinter window is closed.

Answer 1

From your comments it sound's like you do not need a GUI at all. Just write the image to disk and call an external viewer.

On most systems it should be possible to launch the default viewer using something like this:

import subprocess 

subprocess.Popen("yourimage.png")