Python Tkinter Gui Hide and show key bind - python

0 iq Question incoming, i wanna know is there a way i can show and hide a tkinter gui with insert key, ive searched online but wasnt able to find an answer, for example like csgo menus.
Thank you.

Please make sure to always include what you have tried and your research. Follow these guidelines to create a minimal reproducible example.
What you are trying to accomplish can be done with this Tkinter template:
from tkinter import *
root = Tk()
# Open new window
def launch():
global second
second = Toplevel()
second.title("Child Window")
second.geometry("400x400")
# Show the window
def show():
if event.keysym == "Insert"
second.deiconify()
# Hide the window
def hide():
if event.keysym == "Insert"
second.withdraw()
# Add Buttons
Button(root, text="launch Window", command=launch).pack(pady=10)
Button(root, text="Show", command=show).pack(pady=10)
Button(root, text="Hide", command=hide).pack(pady=10)
root.mainloop()

Related

Tkinter winfo_ismapped() method not working

I have a program in python which in which I use Listboxes, buttons and labels. So today I came conflicting with a problem. I wanted to make my listbox appear when a button is clicked and disappear when the same button is clicked again. How can I achieve this? I tried using the winfo_ismapped() method but didnt seem to work. I think I might have done something crazy. If so, please point it out and give me a corrected answer. Else please tell me a better way to do it.
My Code:
import tkinter as tk
from tkinter import *
root = tk.Tk()
root.geometry('500x500')
def showMenu():
overlay = Listbox(root, bg="green", height=22, width=58)
if overlay.winfo_ismapped() == 0:
overlay.place(x=0,y=35)
else:
overlay.placeforget()
button = tk.Button(root,text="place/remove", command=showMenu)
button.place(x=0,y=0)
root.mainloop()
Actually it comes when I press the button but hide after I press it again.
In the same way I have another issue with these labels too.
CODE:
import tkinter as tk
root = tk.Tk()
def placeFun():
successtext = tk.Label(root, text="Success", anchor='nw', bg="#212121", fg="#ff3300",font=("Consolas", 15, "bold"))
if successtext.winfo_ismapped() == 0:
successtext.place(x=0,y=50)
else:
succestext.forget()
button = tk.Button(root, text='place/rem', width=25, command=placeFun)
button.place(x=0,y=0)
root.mainloop()
Please Note: I want a professional way to handle this, I said it because, I know a way in which we use variables like:
globalvartimes = 0
def somefunc():
if times % 2 == 0:
show the listbox
global times
times += 2
else:
remove the listbox
times += 1
*This shows the listbox when times is even and remove it when it's odd.
These makes the code look non-professional and long.
The problem is every time showMenu() is called another Listbox is created. To fix that, create the Listbox outside of the function (so it's a global).
(I also noticed you misspelled the name of place_forget() method.)
import tkinter as tk
from tkinter import *
root = tk.Tk()
root.geometry('500x500')
def showMenu():
if overlay.winfo_ismapped(): # Placed?
overlay.place_forget()
else:
overlay.place(x=0,y=35)
overlay = Listbox(root, bg="green", height=22, width=58)
button = tk.Button(root,text="place/remove", command=showMenu)
button.place(x=0,y=0)
root.mainloop()
This looks like it is what is wrong with your Label example, too.
Note: If you want to write "professional" code, I suggest you read (and start following) the
PEP 8 - Style Guide for Python Code.

How can I create MacOS' Help button in tkinter?

The Human Interface Guidelines describe a distinctive Help button that isn't the same design as a normal tkinter button. My question is simply, can I create this button in tkinter? I looked in the tkinter docs on effbot, but couldn't find anything.
You can make a button, put a question mark image on it, and set borders to 0.
Here is an example:
from tkinter import *
root = Tk()
root.config(bg="light grey")
helpim = PhotoImage(file="help.gif")
help = Button(root, bd=0, bg="light grey")
help["activebackground"] = "light grey"
help.config(image=helpim)
help.pack()
root.mainloop()
This is the question mark image I used:
You can assign the button a command to show the help as a dialog message box.
Output:

How can I combine this tkinter label and string into one window?

In a program I am working on there is a tkinter label/button which starts the card game (the program I am using) and a other window that has a string stating 'Welcome to the card game'.
Here is the tkinter section of the code:
import tkinter
window = tkinter.Tk()
print()
from tkinter import *
def quit():
global root
root.quit()
root = Tk()
while True:
label = tkinter.Label(window, text = "Welcome to the card game! (During name registration only use characters)").pack()
Button(root, text="Start Game", command=quit).pack()
root.mainloop()
However when I run the program they each appear in their own window screens when it would be more convenient for the user to have the options in one single window.
Is there anyway to merge them?
EDIT - (Having the button and text using root has fixed the problem.)
There is a lot going on here that should not be in such a small set of code.
Lets break it down.
First your imports. You are importing from tkinter multiple times. You only need to import once and you can use everything with the proper prefix. The preferred method is import tkinter as tk this way you don't overwrite any other imports or built in methods.
Next we need to get rid of one of your instances of Tk() as tkinter should only ever have one. For other windows use Toplevel().
In your quit function you do not need to define global as you are not assigning values here so the function will look in the global namespace for root.
Next Lets delete the empty print statement.
Next make sure both your label and button have the same container assigned to them. This is the reason why you are seeing them in different windows.
Next rename your function as quit is a built in method and should not be overwritten.
Lastly we remove the while statement as the mainloop() is already looping the Tk instance. You do not need to manage this yourself.
Here is what your code should look like (a 2nd window serves no purpose here):
import tkinter as tk
def root_quit():
root.quit()
root = tk.Tk()
tk.Label(root, text="Welcome to the card game! (During name registration only use characters)").pack()
tk.Button(root, text="Start Game", command=root_quit).pack()
root.mainloop()
Here is an example using Toplevel just so you can get an idea of how it is used.
import tkinter as tk
def root_quit():
root.quit()
def game_window():
top = tk.Toplevel(root)
tk.Button(top, text='exit', command=root_quit).pack()
root = tk.Tk()
tk.Label(root, text="Welcome to the card game! (During name registration only use characters)").pack()
tk.Button(root, text="Start Game", command=game_window).pack()
root.mainloop()

Two text boxes and many buttons python

I would like to press a button and have it print the location of the cursor X/Y coordinates as well as the content of the button in tkinter.
I'm new to python and I haven't found any way to do this. Can anyone help me finding the right start on this issue?
You can use the bind the button click event to print the location clicked, here's an example of how you can do that.
import tkinter as tk
def print_location(event):
print(event.x, event.y)
root = tk.Tk()
button = tk.Button(root, text='click me')
button.bind("<Button-1>", print_location)
button.pack()
root.mainloop()
Note: using external modules such as pyautogui is not recommended and unnecessary.

Keep a menu open in Tkinter

I want to keep a menu cascade open, after a command button within the cascade is clicked. So it basically only closes when the user clicks anywhere else (like it would normally too). Can't seem to find a proper option or a method to open said menu in the callback. The invoke() function only works on buttons wihtin the cascade right? How would you go about that?
Yes, I know this was asked a long time ago, but I was curious if there was any way to accomplish this with tkinter, so I fiddled about for a while and figured out how to do it. I was unable to come up with a way to properly place the persistent menu where it was when it originally opened, but I have managed to make it persist in any location you request (I use upper-left corner of root window). And yes, I know this isn't a nice proper class based implementation, but I was just going for as simple a test as I could write without obscuring it with too many extraneous details.
try:
from tkinter import *
from tkinter.ttk import *
except:
from Tkinter import *
from ttk import *
root = Tk()
var = StringVar()
def menu_click(menu, item):
global root
var.set(item)
menu.post(root.winfo_rootx(), root.winfo_rooty())
root.option_add('*tearOff', False) # remove tearoff from all menus
Label(root, textvariable=var).pack() # just to give menu clicks some feedback
root.geometry('400x300')
menubar = Menu(root)
root['menu'] = menubar
menu_test = Menu(menubar)
menubar.add_cascade(menu=menu_test, label='Test')
menu_test.add_command(label='One', command=lambda: menu_click(menu_test, 'One'))
menu_test.add_command(label='Two', command=lambda: menu_click(menu_test, 'Two'))
menu_test.add_command(label='Three', command=lambda: menu_click(menu_test, 'Three'))
menu_cas = Menu(menu_test)
menu_test.add_cascade(menu=menu_cas, label='Four')
menu_cas.add_command(label='One', command=lambda: menu_click(menu_cas, 'Fourty One'))
menu_cas.add_command(label='Two', command=lambda: menu_click(menu_cas, 'Fourty Two'))
menu_cas.add_command(label='Three', command=lambda: menu_click(menu_cas, 'Fourty Three'))
root.mainloop()

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