I am running tkinter for Python 3 on MacOS and I've run into a problem that I can't solve through google searches. When I call Checkbutton() and add a text parameter, no text is apparent in the window. This is true in even the most sterile of contexts:
from tkinter import *
root = Tk()
test = IntVar()
Checkbutton(root, text = 'test', variable = test).pack()
root.mainloop()
The resulting window looks like this:
I am not sure if I'm just missing something blatantly obvious, but every example I have found online looks more or less like the code above and they have no problems displaying the text. Any help would be greatly appreciated.
Related
I want to create a terminal like design using tkinter. I also want to include terminal like function where once you hit enter, you would not be able to change your previous lines of sentences. Is it even possible to create such UI design using tkinter?
An example of a terminal design may look like this:
According to my research, i have found an answer and link that may help you out
Firstly i would like you to try this code, this code takes the command "ipconfig" and displays the result in a new window, you can modify this code:-
import tkinter
import os
def get_info(arg):
x = Tkinter.StringVar()
x = tfield.get("linestart", "lineend") # gives an error 'bad text index "linestart"'
print (x)
root = tkinter.Tk()
tfield = tkinter.Text(root)
tfield.pack()
for line in os.popen("ipconfig", 'r'):
tfield.insert("end", line)
tfield.bind("<Return>", get_info)
root.mainloop()
And i have found a similar question on quora take a look at this
After asking for additional help by breaking down certain parts, I was able to get a solution from j_4321 post. Link: https://stackoverflow.com/a/63830645/11355351
I have written some code in python for a live time in tkinter.
Whenever I run the code it comes up with some numbers on the tkinter window like 14342816time. Is there a way to fix this?
import tkinter
import datetime
window = tkinter.Tk()
def time():
datetime.datetime.now().time()
datetime.time(17, 3,)
print(datetime.datetime.now().time())
tkinter.Label(window, text = time).pack()
window.mainloop()
After some fixes to your code, I came up with the following, which should at least get you started toward what you want:
import datetime
import tkinter
def get_time():
return datetime.datetime.now().time()
root = tkinter.Tk()
tkinter.Label(root, text = get_time()).pack()
root.mainloop()
The imports are needed so that your program knows about the contents of the datetime and tkinter modules - you may be importing them already, however, I can't tell that for certain from what you posted. You need to create a window into which you put your label, which wasn't happening; following convention, I called that parent (and only) window "root". Then I put the Label into root. I changed the name of your time() function to get_time(), since it's best to avoid confusing fellow programmers (and maybe yourself) with a function that shares its name with another (the time() function in time). I removed two lines in get_time() that don't actually accomplish anything. Finally, I changed the print you had to a return, so that the value can be used by the code calling the function.
There are other improvements possible here. If you're content with the time as it is, you could eliminate the get_time function and just use datetime.datetime.now().time() instead of calling get_time(). However, I suspect you might want to do something to clean up that time before it is displayed, so I left it there. You might want to research the datetime and time modules some more, to see how to clean things up.
I am trying to make a popup window where someone can fill in a string in an Entry box. I have gone through many examples, but it doesn't work.
I am trying to this:
var_entry = simpledialog.askstring("Test", "Test")
I get this error message:
_tkinter.TclError: window ".!_querystring" was deleted before its visibility changed
Thanks in advance!
edit: posted wrong error message
I know this is an old thread, but I'm having the same problem and so far haven't found the root cause.
However, this workaround works for me in case anyone else needs it:
#Create a new temporary "parent"
newWin = Tk()
#But make it invisible
newWin.withdraw()
#Now this works without throwing an exception:
retVal = simpledialog.askstring("Enter Value","Please enter a value",parent=newWin)
#Destroy the temporary "parent"
newWin.destroy()
I was also able to work around the problem by using the above workaround suggested by John D.
I did some research on this, and it seems that this exception is raised when all of the following conditions are met.
The thread that called the simpledialog.askstring method is not the main thread.
The Tk window specified in the parent or the Tk window specified in the default_root variable is different from the thread that called the simpledialog.askstring method.
However, I could not come up with a process to deal with this problem. I hope this helps to solve the problem.
This issue is caused by askstring is not called in the main thread, which is the main thread.
Make sure to call this method in the main thread. For example, this code works well.
from tkinter import Tk
from tkinter.simpledialog import askstring
a = Tk()
askstring('1', '2')
a.mainloop()
And this code will throw your exception when you close the dialog window.
from tkinter import Tk
from tkinter.simpledialog import askstring
from threading import Thread
a = Tk()
Thread(target=askstring, args=('1', '2')).start()
a.mainloop()
I'm not sure without your code example but I figured that this error message means that the variable where you are putting the 'askstring' return value or initialvalue is going out of scope before the dialog window is finished.
When I had this error message I placed a declaration of the variables outside the inner scope. Forgive me if I am talking here using C concepts (which most Python users would rather ignore).
answer = "dummy"
query_str = "dummy" # without these lines query_str and answer
# can be cleaned up by Python
# before the 'askstring' is done with them
while (1):
query_str = "0"
answer = simpledialog.askstring("Get Number",
"Enter NextNumber",
initialvalue = query_str)
print(answer)
print(query_str)
time.sleep(1)
Adding the declarations outside the loop scope worked for me.
So I'm working on a tkinter project and one issue I come across is finding a way to flip/rotate a button object's text vertically. One way I can kinda cheat into making this happen is putting a canvas object on top of the button with the canvas being drawn last (as shown below) but is there a cleaner way to approach this by just manipulating the Button object attributes?
from tkinter import*
root = Tk()
windowDimensions = (1300,600)
root.title("Mapper")
root.geometry(str(windowDimensions[0])+"x"+str(windowDimensions[1]))
button1=Button(root,text='',width=2,height=9)
button1.place(x=0,y=20)
can = Canvas(root,width=15,height=80)
can.place(x=2,y=30)
can.create_text(0, 80, anchor="nw", angle=90,text='hello',font=("Purisa", 12))
root.mainloop()
Edit: A problem I get with doing it this way is any place where the canvas is on the button, it obstructs the ability to click where the canvas is.
Your best option (which isn't a great option) is to screenshot the button, rotate it in an image editor, and then use that image in your button instead of text.
from tkinter import*
root = Tk()
# .gif file encoded as base64
vert_button_data = '''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=
'''
windowDimensions = (1300,600)
root.title("Mapper")
root.geometry("{}x{}".format(*windowDimensions))
button1_image = PhotoImage(data=vert_button_data)
button1=Button(root,image=button1_image)
button1.place(x=0,y=20)
root.mainloop()
You'll lose the hover animation but again that's something you can recreate with images.
To get the base64 encoded data from a .gif you can use this:
import codecs
with open('export.gif', 'rb') as f:
print(codecs.encode(f.read(), 'base64').decode())
The problem I am encountering is that I appear to be stuck in an infinite loop, (If I am not, please correct me). I am using tkinter for python 3.6 (64 bit) on windows 10.
In the module I am having an issue with I have 3 entry widgets and 2 buttons. Both buttons call the "destroy()" function in order to kill the parent window.
Below is a heavily abstracted version of my module, the purpose of the module is to take inputs from the entry widget and write them to a file.
def Create():
parent = Tk()
parent.MakeItlookNice
entry1 = Entry(parent)
entry1.insert(INSERT, "Please enter your desired username here")
entry2 = Entry(parent)
entry2.insert(INSERT, "Please enter your desired password here")
entry3 = Entry(parent)
entry3.insert(INSERT, "What is your mother's maiden name")
Submit = tk.Button(parent,
text ="Click here to submit your stuff",
command = lambda: [parent.destroy(),
submit.function()])
Cancel = tk.Button(parent,
text ="Click here to cancel your request",
command = lambda: parent.destroy())
parent.mainloop()
This function is contained within the module "RegisterNewUser". The "Menu" module is the module that called this function. As far as I am aware once parent.destroy() is called there is no more code to execute since it is all contained within parent.mainloop(), therefore the function is finished and the "Menu" module should continue executing.
What should happen:
I want the Submit button to destroy the window, execute the function and then return to the "Menu" module.
I want the cancel button to destroy the window and return to the "Menu" module.
What actually happens:
The window closes, like it is supposed to
But the code inside the "Menu" module does not start executing again
When I go to close the python shell, it warns me that the program is still running
Ultimately my question is, what code is still running and why hasn't it stopped?
Thank you for reading this and if you require more detail please let me know.
EDIT: I have done research on this topic before posting this question. I have read the documentation on both the tk.destroy() function and the tk.mainloop() function, I have also opened up the Tkinter module in IDLE to try and understand what happens at a deeper level but after all this, I was still unable to figure out a solution. This is my first question on stack overflow, please forgive me if I have done anything wrong.
Hmmm, so you say multiple windows? an easier way to achieve a complex UI as such is using a concept called frames. Tkinter allows you to completely change you screen and layout if you switch to a new frames. This might require you to reprogram alot of code. for an example see Switch between two frames in tkinter
Also, Some guy built a really nice Bitcoin monitoring app using tkinter and frames on youtube
I think you would probably benefit from using Toplevel() here.
I have taken the code you provided and added it to a class used to create the main window and to manage the pop up window.
I noticed a few things with you code.
Its obvoious you are importing tkinter twice like this:
import tkinter as tk
from tkinter import *
I can tell from how you have written your entry fields vs your buttons. Do not do this. Instead just used one or the other. I recommend just using import tkinter as tk.
You are using a function to create a new tkinter instance and judging by your question you all ready have a tkinter instance created for your menu. Instead of creating new Tk() instances you can use Toplevel() instead to open a new window that can inherit everything from the main window and should be easier to manage.
You do not really need to use lambda in this situation. I have also removed the lambda function and replaced with a simple command that will work here.
Take a look at the below code and let me know if you have any questions.
import tkinter as tk
class MyApp(tk.Frame):
def __init__(self, master, *args, **kwargs):
tk.Frame.__init__(self, master, *args, **kwargs)
self.master = master
self.master.title("Main MENU")
tk.Button(self.master, text="Open top level", command = self.create).pack()
def create(self):
self.top = tk.Toplevel(self.master)
entry1 = tk.Entry(self.top, width = 35)
entry1.pack()
entry1.insert(0, "Please enter your desired username here")
entry2 = tk.Entry(self.top, width = 35)
entry2.pack()
entry2.insert(0, "Please enter your desired password here")
entry3 = tk.Entry(self.top, width = 35)
entry3.pack()
entry3.insert(0, "What is your mother's maiden name")
tk.Button(self.top, text ="Click here to submit your stuff",
command = self.Submit).pack()
tk.Button(self.top, text ="Click here to cancel your request",
command = self.top.destroy).pack()
def Submit(self):
print("Do something here to submit data")
self.top.destroy()
if __name__ == "__main__":
root = tk.Tk()
app1 = MyApp(root)
tk.mainloop()
You can use toplevel() and its library function wait_window() just prior to (or instead of) your mainloop() and your problem will be solved
wait_window() mentioned above worked for me in the code below replacing popup.mainloop() with it, the mainloop() kept my code in an infinite loop upon the popup even after okay button was hit