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)
Related
from PIL import Image, ImageTk
from tkinter import *
from tkinter import Label
def open_window():
menu = Toplevel(root)
menu.geometry("800x800")
menu.title("my game's menu")
menu.resizable(False, False)
menu.geometry("800x800")
lbl = Label(menu, text ="Hello!").pack
menu.mainloop()
root = Tk()
root.geometry("400x300")
Label = Label(root, text="Are you ready?")
Label.pack()
root.title("quick question")
btn = Button(root, text="Yes", command= open_window)
btn.pack(padx=20, pady = 20)
root.mainloop()
I got this error while I was working on previously seen code: File "C:\Users\User\Desktop\naujas zaidimas\scratch.py", line 11, in open_window
lbl = Label(menu, text ="Hello!").pack
TypeError: 'Label' object is not callable
Does anyone know why/how to fix it?
The code has a number of fairly obvious errors and one insidious error
This import plays no role in current code
from PIL import Image, ImageTk
Not a good or preferred way to import tkinter
This will cause problems later on
from tkinter import *
This is unnecessary with the current inport method
from tkinter import Label
def open_window():
The rule for Python functions is: Names created in functions stay in functions.
It will require a global instruction to make 'menu' available elsewhere in your code
menu = Toplevel(root)
menu.geometry("800x800")
menu.title("my game's menu")
menu.resizable(False, False)
Duplicated geometry instruction
menu.geometry("800x800")
This object has already been defined as a Label object so trying to name it throws a TypeError
lbl = Label(menu, text ="Hello!").pack()
This is unnecessary since root.mainloop() has already been executed
menu.mainloop()
root = Tk()
root.geometry("400x300")
Here is another naming problem caused by the import method chosen
Label = Label(root, text="Are you ready?")
Label.pack()
root.title("quick question")
This button will enable you to create MANY Toplevel windows
The problem is, ALL of them will be called 'menu'!?
This is the insidious error
btn = Button(root, text="Yes", command= open_window)
btn.pack(padx=20, pady = 20)
root.mainloop()
# This solution avoids all the previous problems
import tkinter as tk
root = tk.Tk()
root.title("quick question")
root.geometry("400x300")
tk.Label(root, text = "Are you ready?").pack()
# This will give tkinter time to process the given instructions.
root.update()
menu = tk.Toplevel(root)
# withdraw will make menu temporarily invisible
menu.withdraw()
menu.title("my game's menu")
menu.geometry("800x800")
menu.resizable(False, False)
tk.Label(menu, text = "Hello!").pack()
# command will now make menu window visible
btn = tk.Button(root, text = "Yes", command = menu.deiconify)
btn.pack(padx = 20, pady = 20)
root.mainloop()
from PIL import Image, ImageTk
from tkinter import *
def open_window():
menu = Toplevel(root)
menu.geometry("800x800")
menu.title("my game's menu")
menu.resizable(False, False)
menu.geometry("800x800")
lbl = Label(menu, text ="Hello!").pack()
menu.mainloop()
root = Tk()
root.geometry("400x300")
lbl1 = Label(root, text="Are you ready?").pack()
root.title("quick question")
btn = Button(root, text="Yes", command= open_window)
btn.pack(padx=20, pady = 20)
root.mainloop()
I Played around both our codes and now somewhy it works.
from tkinter import *
from tkinter import ttk
root = Tk()
root.mainloop()
button1 = ttk.Label(root, text = 'lol')
button1.pack()
And when i try to run the program only the window show up but without the button.
it gives me this error:
Exception has occurred: TclError
NULL main window
File "C:\Users\Elad\Desktop\coding\tkintertut.py", line 6, in <module>
button1 = ttk.Label(root, text = 'lol')
The root.mainloop() has to be at the end of the code for the window to be shown, not right after declaration of root.
from tkinter import *
from tkinter import ttk
root = Tk()
button1 = ttk.Label(root, text = 'lol')
button1.pack()
root.mainloop()
Explanation:
Only the lines of code between root and root.mainloop() gets executed as long as the window is open, if you close the window, the Labels and all other widget declared after root.mainloop() become active but, the window is now closed and root is destroyed, hence the error. But here in my answer, the code between root and root.mainloop() has everything I need to be in the window and hence it shows the complete window. Just keep in mind to always say root.mainloop() at the end of the code only
Hope it cleared your errors. Do let me know if any more errors or doubts.
Cheers
Your root.mainloop() should be the last code!
from tkinter import *
from tkinter import ttk
root = Tk()
button1 = ttk.Label(root, text = 'lol')
button1.pack()
root.mainloop()
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.
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()
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()