First of all note that I am doing all these in Enthought/Canopy.
I have this basic GUI where it has 4 entry boxes in which you type stuff, and then you hit a button and it prints the stuff you entered. However, it is not operating the way I want it to. When you start the code and type in values and press the button, it won't do anything, then you close the gui window and it will print the values you entered.
Also, when you entered the values and hit the button once, again it won't do anything, but if you hit the button again it will print the values but not as expected. Say you entered 1, 2, 3, 4 and hit the button twice, the thing you see on the screen is 1, 2, 3, 4, 1 and when you close the window now, it will print out the rest.
I'd appreciate if you people can help me out with this. Thank you in advance. Below is my code:
Update: This issue does not happen with IDLE, but only Canopy.
from Tkinter import *
class Application:
def printcmd(self):
print(self.entrybox.get())
def __init__(self, master):
self.entrybox = Entry(master)
self.button = Button(master, text="print", command = self.printcmd)
self.entrybox.grid()
self.button.grid()
root = Tk()
Application(root)
root.mainloop()
If you are doing this in the Canopy GUI, you should ensure that Qt is not already set as the GUI backend. See https://support.enthought.com/hc/en-us/articles/204469880-Using-Tkinter-Turtle-or-Pyglet-in-Canopy-s-IPython-panel
For the print lag: Python buffers its output. If you want to ensure that some output is printed immediately, follow the print statement with sys.stdout.flush() to flush the print output buffers. (Of course you must first import sys.)
This can be an issue in any Python program. It arises more frequently in Canopy than in IDLE because Canopy used IPython's QtConsole which separates the execution kernel from the front end terminal-like panel into two separate OS processes.
Related
I am looking for a simple way to display changing real time data in a GUI in python. I am connected to 2 devices and want to display data constantly (like 20 different values), and when I press a button I want to control the one device.
Unfortunately I fail already with the display of the data. For this I have looked at some tkinter tutorials and explanations.
My idea was to implement it with a config function and to overwrite the label continuously. As example how I wanted to display one value:
import tkinter as tk
from pydualsense import pydualsense
# connect to the device
dualsense = pydualsense()
dualsense.init()
# create a window
window = tk.Tk()
# function for updating data
def show_data():
global dualsense
data_label_output.config(text=dualsense.state.LX)
# showing the data as a lable
data_label_output = tk.Label(window)
data_label_output.grid(row=1, column=1)
show_data()
#### or different solution
# showing the data as a lable
data_label_output = tk.Label(window, comand=show_data)
data_label_output.grid(row=1, column=1)
window.mainloop()
Unfortunately, the value is displayed only once at the beginning and nothing changes after that.
Another problem:
When I press the button, I want to be able to control the one device. For this I have a while True loop that permanently checks if a button is pressed and then executes actions. As a separate program no problem, but how do I integrate this into the tkinter GUI? When I start this PyCharm always crashes.
I use PyCharm and Python 3.8
About simple and functional ideas I would be happy, also to other tools/modules etc., as long as you can easily and quickly implement the idea. It's only for a research project and the programming is only a means to an end.
You can use the after method in tkinter to run something after a short delay. The following code will run show_data once the GUI is ready and then again every 1000 milliseconds.
import tkinter as tk
from pydualsense import pydualsense
# connect to the device
dualsense = pydualsense()
dualsense.init()
# create a window
window = tk.Tk()
# function for updating data
def show_data():
global dualsense
data_label_output.config(text=dualsense.state.LX)
window.after(1000,show_data)
# showing the data as a lable
data_label_output = tk.Label(window)
data_label_output.grid(row=1, column=1)
window.after_idle(show_data)
window.mainloop()
This resolves the updating issue, I'm not sure what behaviour you want when you press the button but if you elaborate and explain, I might be able to help and update this answer.
I work with Python 3.5 and TKinter.
I defined a label and file dialog that updates this label.
A button is responsible to launch this dialog.
self.sel_folder_val = the label that will be updated.
The code:
self.sel_folder_val['text']=filedialog.askdirectory()
After pressing the button in order to launch this dialog, the button stays pressed. Any dialog that a button is responsible to open cause the button to stay low (pressed) after closing this dialog.
I have tried this also with no help...:
self.select_folder_btn.config(relief=RAISED)
Code example:
self.select_folder_btn = Button(self.top)
self.select_folder_btn.place(relx=0.07, rely=0.57, height=34, width=187)
self.select_folder_btn.configure(activebackground="#d9d9d9")
self.select_folder_btn.configure(activeforeground="#000000")
self.select_folder_btn.configure(background="#d9d9d9")
self.select_folder_btn.configure(disabledforeground="#a3a3a3")
self.select_folder_btn.configure(font=self.font3)
self.select_folder_btn.configure(foreground="#000000")
self.select_folder_btn.configure(highlightbackground="#d9d9d9")
self.select_folder_btn.configure(highlightcolor="black")
self.select_folder_btn.configure(pady="0")
self.select_folder_btn.configure(text='''Select destination folder''')
self.select_folder_btn.bind('<Button-1>',self.update_folder_value)
def update_folder_value(self,event):
self.sel_folder_val['text']=filedialog.askdirectory()
return
After executing update_folder_value() function, self.select_folder_btn stays down.
I used the command:
self.select_folder_btn.configure(command=self.update_folder_value)
Instead of bind:
self.select_folder_btn.bind('<Button-1>',self.update_folder_value)
It solved my problem.
Thanks
First for future reference this is a minimal working example:
from Tkinter import *
import tkFileDialog as filedialog
class app:
def __init__(self):
self.top = Tk()
self.select_folder_btn = Button(self.top)
self.select_folder_btn.place(relx=0.07, rely=0.57, height=34, width=187)
self.select_folder_btn.configure(activebackground="#d9d9d9")
self.select_folder_btn.configure(activeforeground="#000000")
self.select_folder_btn.configure(background="#d9d9d9")
self.select_folder_btn.configure(disabledforeground="#a3a3a3")
#self.select_folder_btn.configure(font=self.font3)
self.select_folder_btn.configure(foreground="#000000")
self.select_folder_btn.configure(highlightbackground="#d9d9d9")
self.select_folder_btn.configure(highlightcolor="black")
self.select_folder_btn.configure(pady="0")
self.select_folder_btn.configure(text='''Select destination folder''')
self.select_folder_btn.configure(command=self.update_folder_value)
self.sel_folder_val = {}
self.top.mainloop()
def update_folder_value(self):
self.sel_folder_val['text']=filedialog.askdirectory()
self.top.update_idletasks()
app()
and even that's not minimal. Second your problem is hard to find since this isn't minimal- you're doing something really weird - binding the button to a click. You're overriding the built-in binding, and apparently it still affects the state of the button on press, but not going back. What you wanted is:
self.select_folder_btn.configure(command=self.update_folder_value)
instead of your:
self.select_folder_btn.bind('<Button-1>',self.update_folder_value)
You could also define that in the Button command. What you did is bypassed the button mechanism, so apparently only half of it is executed, and the relief is not raised. Note you have to remove the event parameter your method accepts.
Part of my code is as follows:
def get_songs():
label6.configure(text='Wait')
os.system('/home/norman/my-startups/grabsongs')
label6.configure(text='Done')
The label is not updated at the first .configure() but is at the second one.
Except if I cause a deliberate error immediately after the first one at which point it is updated and then the program terminates.
The system call takes about 2 minutes to complete so it isn't as if there isn't time to display the first one.
I am using Python 2.7.6
Does anyone know why please?
I'm going to guess you're using Tkinter. If so, as #albert just suggested, you'll want to call label.update_idletasks() or label.update() to tell Tkinter to refresh the display.
As a very crude example to reproduce your problem, let's make a program that will:
Wait 1 second
Do something (sleep for 2 seconds) and update the text to "wait"
Display "done" afterwards
For example:
import Tkinter as tk
import time
root = tk.Tk()
label = tk.Label(root, text='Not waiting yet')
label.pack()
def do_stuff():
label.configure(text='Wait')
time.sleep(2)
label.configure(text='Done')
label.after(1000, do_stuff)
tk.mainloop()
Notice that "Wait" will never be displayed.
To fix that, let's call update_idletasks() after initially setting the text:
import Tkinter as tk
import time
root = tk.Tk()
label = tk.Label(root, text='Not waiting yet')
label.pack()
def do_stuff():
label.configure(text='Wait')
label.update_idletasks()
time.sleep(2)
label.configure(text='Done')
label.after(1000, do_stuff)
tk.mainloop()
As far as why this happens, it actually is because Tkinter doesn't have time to update the label.
Calling configure doesn't automatically force a refresh of the display, it just queues one the next time things are idle. Because you immediately call something that will halt execution of the mainloop (calling an executable and forcing python to halt until it finishes), Tkinter never gets a chance to process the changes to the label.
Notice that while the gui displays "Wait" (while your process/sleep is running) it won't respond to resizing, etc. Python has halted execution until the other process finishes running.
To get around this, consider using subprocess.Popen (or something similar) instead of os.system. You'll then need to perodically poll the returned pipe to see if the subprocess has finished.
As an example (I'm also moving this into a class to keep the scoping from getting excessively confusing):
import Tkinter as tk
import subprocess
class Application(object):
def __init__(self, parent):
self.parent = parent
self.label = tk.Label(parent, text='Not waiting yet')
self.label.pack()
self.parent.after(1000, self.do_stuff)
def do_stuff(self):
self.label.configure(text='Wait')
self._pipe = subprocess.Popen(['/bin/sleep', '2'])
self.poll()
def poll(self):
if self._pipe.poll() is None:
self.label.after(100, self.poll)
else:
self.label.configure(text='Done')
root = tk.Tk()
app = Application(root)
tk.mainloop()
The key difference here is that we can resize/move/interact with the window while we're waiting for the external process to finish. Also note that we never needed to call update_idletasks/update, as Tkinter now does have idle time to update the display.
I am creating a little time management tool, using Tkinter, so I can keep on task at work. I am having trouble with one aspect that I cannot seem to get working. I'm using the error box so that it is displayed in front of all other windows.
As it is now, the program starts a new thread on a function that keeps track of time, and compares it to the time the user entered for their task. Once real time > the time entered by the user, it starts another thread to spawn the tkMessageBox. I have tried this without starting a new thread to spawn the tkMessageBox, and the problem is the same. If the user enters the same time for 2 separate tasks, the error pop up freezes. I'm having trouble finding information on this topic specifically... The behaviour is odd because if I have 2 alerts, lets say 1 at 0600 and one at 0601, but I do not close the first error box that pops up and let it stay up until the second alert triggers, the second alert will just replace the first one(I would like multiple error boxes to pop up if possible). It's only the alerts that have the same trigger time that cause the pop up to freeze though.
This is my first GUI program and only started learning the concept of threading, and GUIs in the past 24 hours, so I'm not sure if this is a problem with threading or the tkMessageBox. Because of the behaviour of the error box, I’m thinking it is the thread module combined with the tkMessageBox module. The command I'm using is:
tkMessageBox.showerror('TIMER ALERT!!!', comp_msg)
Here is the source I put comments in there to help. The tkMessageBox I’m talking about is line 56.
I guess I'm not sure if I can even do what I am trying to do with the pop-up box, if I can, I'm not sure how. If I can't, is there a alternative way to spawn multiple error type pop-up boxes with Tkinter? I just want multiple boxes to be able to appear at any given time.
Thanks in advance, and I really appreciate any help at all.
EDIT:
import thread
from Tkinter import *
#Spawns Error Box. Runs in it's own thread.
def message_box(comp_msg,q): # q is an empty string because of thread module.
print "Spawning Error Box..."
eb =Toplevel()
eb.config(master=None,bg="red")
pop_l = Label(eb,text="ALERT!!!")
pop_l2=Label(eb,text=comp_msg)
pop_l.pack(pady=10,padx=10)
pop_l2.pack(pady=15,padx=10)
return eb
thread.start_new_thread(message_box,(comp_msg,""))
tkmessageBox default dialog boxes are modal. You could implement a simple none modal dialog box for this application. Here is a good document about creating custom dialog boxes.
This way you can create as many new custom dialog boxes as your app requires, since each one is just a new Toplevel.
Here is a simple Tkinter app that shows the clock on the main window. When you click on the button it starts new tkMessageBox dialog boxes in new threads. (If you run it) You could see that the main thread that runs the TK event loop is working (since the time is getting updated), but the error boxes are not showing up as expected.
#!/usr/bin/env python
import datetime
import threading
from Tkinter import *
import tkMessageBox
class MyApp(Frame):
def __init__(self, root=None):
if not root:
root = Tk()
self.time_var = StringVar()
self.time_var.set('starting timer ...')
self.root = root
Frame.__init__(self, root)
self.init_widgets()
self.update_time()
def init_widgets(self):
self.label = Label(self.root, textvariable=self.time_var)
self.label.pack()
self.btn = Button(self.root, text='show error', command=self.spawn_errors)
self.btn.pack()
def update_time(self):
self.time_var.set( str(datetime.datetime.now()) )
self.root.after(1000, self.update_time)
def spawn_errors(self):
for i in range(3):
t = threading.Thread(target=self.show_error)
t.start()
def show_error(self):
now = datetime.datetime.now()
tkMessageBox.showerror('Error: %s' % (str(now)), now)
if __name__ == '__main__':
app = MyApp()
app.mainloop()
I am able to get a frame to open from a button. I can close the frame and reopen it from the same button but it throws an error everytime I push the button.
What is throwing the error in my code is root.Show() , it gives me a AttributeError Show error
My question is, although it is working beautifully, could it develop a serious problem for my application?
EDIT: This is the code in my python file
from Tkinter import *
root = Tk()
root.title("Help")
scrollbar = Scrollbar(root)
scrollbar.pack(side=RIGHT, fill=Y)
help_message = 'This is the help menu. Please scroll through the menu to find the answer to your question'
txt = Text(root, wrap=WORD) # wrap=CHAR, wrap=NONE
txt.pack(expand=1, fill=BOTH)
txt.insert(END, help_message)
txt.config(yscrollcommand=scrollbar.set, state=DISABLED)
scrollbar.config(command=txt.yview)
root.Show()
It is taking the error from the last line of this program. And this is the error in the command prompt:
You probably just want to use root.mainloop(), because I don't see anywhere the method Show or show (Python is case sensitive, and methods/functions are usually lower_case_with_underscores).
The mainloop function essentially waits for the program to end, but if you have things like buttons, you can have Tkinter call a certain function to respond (event driven).
I removed root.Show(), added a button that has to be clicked in order for the window to stay hidden but running. I then added in the main file HelpBox.root.deiconify() which makes the screen pop up. If the user does click the "X" button, then the help menu cannot be reopened until they restart the application.
Thanks for everyone's help and ideas