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.
Related
Is it possible to have two instances of tkinter ?
import tkinter as tk
import tkinter as sk
root = tk.Tk()
root2 = sk.Tk()
....some window with tk
....some window with sk
root.mainloop()
root2.mainloop()
then have a Toplevel() in both instances.
You can, but the way it works likely won't be like what you expect. Importing it twice isn't the problem (but neither is it the solution). No matter how you import it or how often you import it, creating more than one instance of Tk is the problem. Each instance is backed by a separate internal interpreter. Images and variables and widgets created in one won't exist in the other.
If you need more than one window, it's usually best if second and subsequent windows are instances of Toplevel.
By creating 2 instances or tkinter I can close one instance and all top Levels of that instance. Leaving the second Instance open and running with it's Toplevel instances. This is useful in the sense where I use it for a User preference file. Where it checks for a userfile database for keeping track of variables. I use a csv file to save user variable using pandas. This in turn keeps all the form information in my app safe from being erased after closing the application or accidentally closing the window. Adding AtExit saves the info closing the addinfo window and continues to run the main application. That was my reasoning for having asked the question. I have since found that using multiple Toplevel(s) is a better choice as it will also produce the same result. So my menu items all have a separate instance definition and can be closed in the same manner ,making error checks with each closed window.
from tkinter import Tk,Label,Entry,Button,Toplevel
root=Tk()
def about():
ab=Toplevel()
# About stuff for this window
ab.mainloop()
def info():
inf=Toplevel()
# Information Stuff for this window.
inf.mainloop()
def getUserInfo():
# User info using pandas
getUserInfo.mainloop()
root.mainloop()
I've seen a couple questions similar to this, but they both appear to involve VBA and not Python.
This is a relatively recent error, so I suspect it might have something to do with the fact that I'm using Python 3.7 now.
Basically, using the Dispatch method from win32com.client, I am able to open a new Excel workbook and make my edits as I always have been. However, for some reason, I am unable to tell the application to quit successfully.
Used to be that I could write:
self.excel_app.Quit()
But now, I'm getting an AttributeError, of all things. Saying the Excel.Application does not have a Quit() attribute. Again, this is Python 3.7. What happened?
[EDIT]
Relevant code:
import sys
from win32com.client import Dispatch
#...
class X(object):
def __init__(self):
#...
self.excel_app = Dispatch("Excel.Application")
self.report_workbook = self.excel_app.Workbooks.Add()
#...
def close_excel(self):
try:
self.excel_app.Quit()
except Exception as ex:
sys.stdout.write("Could not quit application.\n-> ({}) {}\n".format(ex.__class__.__name__, ex))
self.excel_app = None
The exception printed to the terminal is:
Could not quit application.
(AttributeError) Excel.Application.Quit
Turns out that my error was coming from the fact that I was still using threading.Thread as the base class for the object trying to do this. Apparently it doesn't work so well with the Dispatch Excel objects anymore, and I'm trying to move away from Threads anyway.
import sys
import webbrowser
import hou
from PySide2 import QtCore, QtUiTools, QtWidgets, QtGui
# Calling UI File & Some Modification
class someWidget(QtWidgets.QWidget):
def __init__(self):
super(someWidget,self).__init__()
ui_file = 'C:/Users/XY_Ab/Documents/houdini18.5/Folder_CGI/someUI.ui'
self.ui = QtUiTools.QUiLoader().load(ui_file, parentWidget=self)
self.setParent(hou.qt.mainWindow(), QtCore.Qt.Window)
self.setFixedSize(437, 42)
self.setWindowTitle("Requesting For Help")
window_C = someWidget()
window_C.show()
So, I have created this small script that shows the UI, I have connected this to Houdini Menu Bar. Now The Problem is if I click the menu item multiple times it will create another instance of the same UI & the previous one stays back, What I want is something called "If Window Exist Delete It, Crate New One" sort of thing.
Can someone guide me? I am fairly new to python in Houdini and Qt so a little explanation will be hugely helpful. Also, why can't I use from PySide6 import?? Why do I have to use from PySide2?? Because otherwise Houdini is throwing errors.
For the same thing what used to do in maya is
# Check To See If Window Exists
if cmds.window(winID, exists=True):
cmds.deleteUI(winID)
Trying to do the same thing inside Houdini.
I don't have Maya or Houdini, so I can't help you too much.
According to https://www.sidefx.com/docs/houdini/hom/cb/qt.html
It looks like you can access Houdini's main window. The main reason the window is duplicated or deleted is how python retains the reference to window_C. You might be able to retain the reference to just show the same widget over and over again by accessing the main Houdini window.
In the examples below we are using references a different way. You probably do not need your code that has
self.setParent(hou.qt.mainWindow(), QtCore.Qt.Window)
Create the widget once and keep showing the same widget over and over.
import hou
# Create the widget class
class someWidget(QtWidgets.QWidget):
def __init__(self, parent=None, flags=QtCore.Qt.Window): # Note: added parent as an option
super(someWidget,self).__init__(parent, flags)
...
MAIN_WINDOW = hou.ui.mainQtWindow()
try:
MAIN_WINDOW.window_C.show()
except AttributeError:
# Widget has not been created yet!
# Save the widget reference to an object that will always exist and is accessible
# parent shouldn't really matter, because we are saving the reference to an object
# that will exist the life of the application
MAIN_WINDOW.window_C = someWidget(parent=MAIN_WINDOW)
MAIN_WINDOW.window_C.show()
To delete the previous window and create a new window.
import hou
# Create the widget class
class someWidget(QtWidgets.QWidget):
def __init__(self, parent=None, flags=QtCore.Qt.Window): # Note: added parent as an option
super(someWidget,self).__init__(parent, flags)
...
MAIN_WINDOW = hou.ui.mainQtWindow()
# Hide the previous window
try:
MAIN_WINDOW.window_C.close()
MAIN_WINDOW.window_C.deleteLater() # This is needed if you parent the widget
except AttributeError:
pass
# Create the new Widget and override the previous widget's reference
# Python's garbage collection should automatically delete the previous widget.
# You do not need to have a parent!
# If you do have a parent then deleteLater above is needed!
MAIN_WINDOW.window_C = someWidget() # Note: We do not parent this widget!
MAIN_WINDOW.window_C.show()
Another resource shows you can access the previous widget from the page level variable. https://echopraxia.co.uk/blog/pyqt-in-houdinimaya-basic This is possible, but seems odd to me. The module should only be imported once, so the page level variable "my_window" should never exist. However, it sounds like the Houdini plugin system either reloads the python script or re-runs the import. If that is the case every time you show a new window from the import of the script, you are creating a new window. If the previous window is not closed and deleted properly, Houdini could have an ever growing memory issue.
try:
my_window.close()
except (NameError, Exception):
pass # Normal python would always throw a NameError, because my_window is never defined
my_window = MyWindow()
#This is optional you can resize the window if youd like.
my_window.resize(471,577)
my_window.show()
PySide6
https://www.sidefx.com/docs/houdini/hom/cb/qt.html
The bottom of the page shows how to use PyQt5. The same would apply for PySide6. Houdini just happens to come with PySide2.
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.
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