So you know how when you use Notepad (on Windows), for example, and you want to open an old file? You click file, then open, then a file dialog opens ups and you can select the file you want, and the program will display its contents.
Basically, I want to make a button in Python that can do that exact thing.
Here's my function for the button-
def UploadAction():
#What to do when the Upload button is pressed
from tkinter import filedialog
When I click on the button assigned to this action, nothing happens, no errors, no crash, just nothing.
import tkinter as tk
from tkinter import filedialog
def UploadAction(event=None):
filename = filedialog.askopenfilename()
print('Selected:', filename)
root = tk.Tk()
button = tk.Button(root, text='Open', command=UploadAction)
button.pack()
root.mainloop()
Related
I'm writing a tkinter app that involves a lot of user input fields. One of these fields is a full file path, which is just begging for user error. In order to mitigate this, I'd like to program that particular Entry widget to call filedialog.askopenfilename() when clicked on, so that the user can just pick the file rather than entering the full path manually. As I understand it, the following code should do just that:
# Minimalized version of my code that reproduces the problem
import tkinter as tk
from tkinter import ttk, filedialog
parent = tk.Tk()
val = tk.StringVar()
def get_file(_):
f = filedialog.askopenfilename()
val.set(f)
parent.focus()
label = ttk.Label(parent, text="Put your file here")
label.pack()
enter = ttk.Entry(parent, textvariable=val)
enter.pack()
enter.bind("<FocusIn>", get_file)
parent.mainloop()
When I run this and click on the Entry widget, get_file() gets called twice in a row. The second call comes immediately after I close the first file dialog and overwrites the first file I chose.
Why is this happening, and how do I stop it?
Thanks!
I want to create a button that opens the special directory like "C:\Users\", But as I searched, I did not find any code except the following code:
from tkinter import *
from tkinter import ttk
from tkinter import filedialog
gui = Tk()
gui.geometry("100x100")
def getFolderPath():
filedialog.askdirectory()
btnFind = ttk.Button(gui, text="Open Folder",command=getFolderPath)
btnFind.grid(row=0,column=2)
gui.mainloop()
What is the correct way to solve this problem?
Use os.startfile("C:\\Users") to use the system file explorer to open the directory.
I have a tkinter program that has a button. The button's job is to open a text file on my desktop using notepad. Here is the code:
import tkinter as tk
import os
def open_file():
file = os.path.expanduser('~/Desktop/a.txt')
os.system("notepad.exe " + file)
root = tk.Tk()
button = tk.Button(root, text='open', command=open_file)
button.grid()
root.mainloop()
The code runs and the program is working, but the problem is that when I click the button, the text file opens but tkinter window will be not responding until I close the text file. I am using Python 3 and Windows 10.
Any help would be appreciated.
Thanks.
Just use: os.system("start notepad.exe")
When I run this script, two windows appear, one for the file selection and the Tkinter window. How can I change this so that the Tkinter window only opens after a file has been selected? Thanks
def main():
my_file = askopenfilename()
stage1()
def stage1():
master = Tk()
master.mainloop()
The window master does open only after the file dialog closure (try to change its title to check), the first window you see is the parent window of the file dialog. Indeed, the tkinter file dialogs are toplevel windows, so they cannot exist without a parent window. So the first window you see is the parent window of the file dialog.
The parent window can however be hidden using the withdraw method and then restored with deiconify:
from tkinter import Tk
from tkinter.filedialog import askopenfilename
def main():
master = Tk()
master.withdraw() # hide window
my_file = askopenfilename(parent=master)
master.deiconify() # show window
master.mainloop()
if __name__ == '__main__':
main()
So I have a program that is basically supposed to have a button that opens a file dialog in the (username) folder. But when I run the program it opens without even pushing the button. What's more, the button doesn't even show up. So in addition to that problem I have to find a way to turn the selected directory into a string.
import tkinter
import tkinter.filedialog
import getpass
gui = tkinter.Tk()
user = getpass.getuser()
tkinter.Button(gui, command=tkinter.filedialog.askopenfilename(initialdir='C:/Users/%s' % user)).pack()
gui.mainloop()
Regarding your first issue, you need to put the call to tkinter.filedialog.askopenfilename in a function so that it isn't run on startup. I actually just answered a question about this this morning, so you can look here for the answer.
Regarding your second issue, the button isn't showing up because you never placed it on the window. You can use the grid method for this:
button = tkinter.Button(gui, command=lambda: tkinter.filedialog.askopenfilename(initialdir='C:/Users/%s' % user))
button.grid()
All in all, your code should be like this:
import tkinter
import tkinter.filedialog
import getpass
gui = tkinter.Tk()
user = getpass.getuser()
button = tkinter.Button(gui, command=lambda: tkinter.filedialog.askopenfilename(initialdir='C:/Users/%s' % user))
button.grid()
gui.mainloop()
You forgot to use a geometry manager on the button:
button = tkinter.Button(window, command=test)
button.pack()
If you don't do it, the button won't be drawn. You might find this link useful: http://effbot.org/tkinterbook/pack.htm.
Note that to pass the command to handler you have to write only the name of the function (Like it's descibed in the other answer).
This is an old question, but I just wanted to add an alternate method of preventing Tkinter from running methods at start-up. You can use Python's functools.partial (doc):
import tkinter
import tkinter.filedialog
import getpass
import functools
gui = tkinter.Tk()
user = getpass.getuser()
button = tkinter.Button(
gui,
command=functools.partial(
tkinter.filedialog.askopenfilename,
initialdir='C:/Users/%s' % user)
)
button.grid()
gui.mainloop()