CODEWITHSUNDEEP
pythontkinter
user3817491
user3817491

Reputation: 117

printing variable from askopenfilename

whenever i print this

def callback():
    file_name = askopenfilename(filetypes=(("Music File", "*.mp3"),
                                      ("Video files", "*.mpeg")))
    file_name = file_name.split("/")[-1]
    a = str(file_name)
    return a 

print a

file_name = Button(root, text="open", command=callback)
file_name.pack(side=Tkinter.TOP)

the output will be like this 

.35430920L

but when i use global, it does print out the file name which i want

def callback():
    global a
    file_name = askopenfilename(filetypes=(("Music File", "*.mp3"),
                                      ("Video files", "*.mpeg")))
    file_name = file_name.split("/")[-1]
    a = str(file_name)
    print a

file_name = Button(root, text="open", command=callback)
file_name.pack(side=Tkinter.TOP)

the output will be like this 

"example.mp3"

Upvotes: 1

Views: 587

Answers (1)

Antti Haapala -- Слава Україні
Antti Haapala -- Слава Україні

Reputation: 133919

The output happens in execution order. In your first example, the variable a is given the value .35430920L before the print a runs, and chronologically the print statement occurs before the file dialog is actually even opened, thus it is impossible to print the value of chosen file there.

In the latter case, being in the callback function, the result is already known. Notice that you do not even need a global variable for that case. (you can remove the global a, unless you need the result outside of that function).

Tkinter uses event-driven programming paradigm, it calls callback functions for each occurring event. What you want to do for any larger program, is to wrap your application and widgets in a class, and store things like filenames in class attributes.

Upvotes: 2

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