I have been developing a tkinter project and I have run into a problem where I have two windows open when the button widget is clicked instead of just a singular window frame. There isn't any specific syntactic error with my code, I was just hoping someone could point me in the right direction for structuring my small GUI program correctly.
I would be really appreciate any support even if it was just an example code or a link.
Thanks.
Welcome Page
from tkinter import *
from from External_Menu import *
root = Tk()
root.state('zoomed')
root.title("Leisure Centre")
title = Label(root, text="Leisure Centre", font=("", 26))
title.place(relx=0.5, rely=0.0, anchor=N)
welcome = Button(root, text="Welcome!", font=("", 18), command=menu)
welcome.place(relx=0.5, rely=0.5, anchor=CENTER)
root.mainloop()`
External_Menu
from tkinter import *
def menu():
root = Tk()
root.state('zoomed')
external_menu_lbl = Label(root, text="External Menu", font=("", 26))
external_menu_lbl.pack()
sign_in_button = Button(root, text="Sign In", command=login_page)
sign_in_button.pack()
sign_up_button = Button(root, text="Sign Up", command=sign_up_page)
sign_up_button.pack()
root.mainloop()
Related
Python 3.8, Win 10 is the os, Toplevel widget does not appear to be working with new window not appearing. Any guidance would be appreciated, thanks!
from tkinter import *
root = Tk()
def popup():
top = Toplevel(root)
my_label_top = Label(top, text="This is a Tkinter Popup")
top.mainloop()
my_button = Button(root, text="Popup, click here", command="popup")
my_button.grid(row=0, column=0)
root.mainloop()
Problem:
The only issue here is that the callback command shouldn't be a string.
Solution:
Remove the quotes around popup and the Toplevel window should appear.
Fixed Code:
from tkinter import *
root = Tk()
def popup():
top = Toplevel(root)
my_label_top = Label(top, text="This is a Tkinter Popup")
my_label_top.pack()
my_button = Button(root, text="Popup, click here", command=popup)
my_button.grid(row=0, column=0)
root.mainloop()
Tips:
Using top.mainloop() is not necessary.
You also forgot to pack() the Label(my_label_top)
Currently, I'm attempting to create a simple main menu for a game using Tkinter as a simple GUI since it's a simple RPG game and Python however the image overlays the buttons in all circumstances.
I've tried using some other solutions like placing them or creating a window but I can't find a straight answer of how to make that either.
import tkinter
from tkinter import *
from PIL import ImageTk, Image
(PIL is from when I was using a JPG before.)
root = Tk()
content = ttk.Frame(root)
root.geometry("600x600")
background = ImageTk.PhotoImage(Image.open("bred.png"))
canvas = tkinter.Canvas(root, width=580, height=600)
content.grid(column=0, row=0)
Btn1 = Button(content, text="Play", width=5, height=1)
Btn2 = Button(content, text="Kill me", width=7, height=1, command =
root.quit)
backgroundlabel = tkinter.Label(root, image=background)
backgroundlabel.image = background
backgroundlabel.place(x=0, y=0, relwidth=1, relheight=1)
Btn1.grid(row=1, column=2, padx=(130))
Btn1.columnconfigure(1, weight=1)
Btn1.rowconfigure(1, weight=1)
Btn2.grid(row=1, column=3, pady=(130))
Btn2.columnconfigure(3, weight=1)
Btn2.rowconfigure(1, weight=1)
root.mainloop()
Currently your background's master is set to root while your buttons are set to a frame. The first thing you need to do is set both to the same master, i.e. changing background master to content:
backgroundlabel = tk.Label(content, image=background)
Next you need to deal with the stacking order. You can call widget.lift() to raise the buttons to top:
Btn1.grid(row=1, column=2, padx=(130))
...
Btn1.lift()
Btn2.grid(row=1, column=3, pady=(130))
...
Btn2.lift()
I have a program calculator, the button code opens a new window, but I want the button calculator run in the same window but not two separate, how do I instead run my code in the same windows?
P.S. the code is not mine, it is as an example
from tkinter import *
root = Tk()
root.title("Math Lab")
root.geometry("1400x1000")
heading = Label(root, text = "Welcome to the MATH Lab", font=("Berlin Sans FB", 40, "bold"), fg= "steelblue").pack()
root.configure(background= "powder blue")
def calculator():
w = Tk()
w.geometry("1400x1000")
def evaluate1():
res.configure(text="Answer: " + str(eval(entry.get())))
def evaluate(event):
res.configure(text="Answer: " + str(eval(entry.get())))
but1 = Button(w, text="Enter", width=10, height=3)
but1.place(x=650, y=100)
but1.config(command=evaluate1)
Label(w, text="Your Expression:").pack()
entry = Entry(w)
entry.bind("<Return>", evaluate)
entry.pack()
res = Label(w)
res.pack()
w.mainloop()
but1=Button(root,text="Calculator",width = 10, height = 3)
but1.place(x=100, y=100)
but1.config(command = calculator)
root.mainloop()
Any suggestions?
Here you are requesting a new window:
w = Tk()
Replace it with
w = root
and you are good to go.
(Well, you'll want to tidy up some x,y offsets, for aesthetics.)
from tkinter import *
def printSomething():
inputValue=textBox.get("1.0","end-1c")
res=response(inputValue)
label = Label(root, text=res)
#this creates a new label to the GUI
label.pack()
root = Tk()
button = Button(root, text="Print Me", command=printSomething)
button.pack()
textBox=Text(root, height=2, width=10)
textBox.pack()
root.mainloop()
I have written a python code that returns text. and print that in tkinter label.while i try to execute it shows "None" in label.
It would probably be better to create the label in the global namespace once and then just update the label every time you press the button.
I also recommend using import tkinter as tk vs from tkinter import * as it provides better maintainability as your code grows and you do not end up overwriting built in methods.
I have updated your code and changed a few things to better fit the PEP8 standard.
import tkinter as tk
def print_something():
label.config(text=text_box.get("1.0", "end-1c"))
root = tk.Tk()
tk.Button(root, text="Print Me", command=print_something).pack()
text_box = tk.Text(root, height=2, width=10)
text_box.pack()
label = tk.Label(root)
label.pack()
root.mainloop()
Just changing your line:
res = response(inputValue)
to
res = inputValue
worked for me, creating a new label every time I pressed the button.
In python i have been making a text editor like Microsoft word but i don't know how to make a text entry box for the user to put input. Here is my code! (ps thank you!)
from tkinter import *
import sys
def doNothing():
print("Test")
root = Tk()
root.title("TextEditor")
root.geometry("300x200")
menu = Menu(root)
root.config(menu=menu)
subMenu = Menu(menu)
menu.add_cascade(label="File", menu=subMenu)
subMenu.add_command(label="New Project...", command =doNothing)
subMenu.add_command(label="Save", command=doNothing)
subMenu.add_separator()
editMenu = Menu(menu)
menu.add_cascade(label="Edit", menu=editMenu)
editMenu.add_command(label="Undo",command=doNothing)
root.mainloop()
You can do that like this:
TextArea = Text()
TextArea.pack(expand=YES, fill=BOTH)
If you want a scrollbar with it:
TextArea = Text()
ScrollBar = Scrollbar(root)
ScrollBar.config(command=TextArea.yview)
TextArea.config(yscrollcommand=ScrollBar.set)
ScrollBar.pack(side=RIGHT, fill=Y)
TextArea.pack(expand=YES, fill=BOTH)
Hope this helped, good luck!
It is an old question but currently following is a very good method for scrollable multiline text entry:
ScrolledText(mainwin, width=50, height=5).pack()
Full program:
from tkinter import *
from tkinter.scrolledtext import ScrolledText
mainwin = Tk()
ScrolledText(mainwin, width=50, height=5).pack()
mainwin.mainloop()
Following demo application shows its usage further and comparison with entry box (for python3):
from tkinter import *
from tkinter.scrolledtext import ScrolledText
mainwin = Tk()
Label(mainwin, text="An Entry Box:").grid(row=0, column=0)
ent = Entry(mainwin, width=70); ent.grid(row=0, column=1)
Button(mainwin, text="Print Entry", command=(lambda: print(ent.get()))).grid(row=0, column=2, sticky="EW")
Label(mainwin, text="ScrolledText Box:").grid(row=1, column=0)
st = ScrolledText(mainwin, height=5); st.grid(row=1, column=1)
Button(mainwin, text="Print Text", command=(lambda: print(st.get(1.0, END)))).grid(row=1, column=2, sticky="EW")
Button(mainwin, text="Exit", command=sys.exit).grid(row=2, column=0, columnspan=3, sticky="EW")
mainwin.mainloop()