How to make restart python program button using tkinter button - user-interface

im try to make restart button for my python program gui My gui but im still confused how to do it. Anyone know how to make the button to restart python code ? I appreciate every help.

use the code below:
import sys
import os
from tkinter import Tk, Label, Button
def restart_me():
pth = sys.executable
os.execl(pth, pth, *sys.argv)
root = Tk()
Label(root, text="hi there!").pack()
Button(root, text="Restart", command=restart_me).pack()
root.mainloop()

Related

Spyder never "quits"

I loaded up Anaconda and realized there is more stuff there than I can absorb in my lifetime. However, I did load Spyder and pasted in a working program that runs without error in PyCharm and Thonny.
"""Create a 2 button window"""
import tkinter as tk
from tkinter import *
from tkinter import messagebox
#write slogan out in a message box
def write_slogan():
messagebox.showinfo("our message",
"tkinter is easy to use")
#set up the window
root = tk.Tk() #get the window
root.geometry("100x100+300+300") #x, y window size and position
#create Hello button
slogan = Button(root,
text="Hello",
command=write_slogan)
slogan.pack(side=LEFT, padx=10)
#create exit button with red letters
button = Button(root,
text="QUIT",
fg="red",
command=quit)
button.pack(side=RIGHT, padx=10)
#start running the tkinter loop
root.mainloop()
It may not surprise Spyder mavens, but I was surprised to see that it did not recognize "quit" However, I replaced it with
command=sys.exit
and that at least compiled.
However, when I ran it, the window appeared UNDER the Spyder window. I found it in the task bar. BUT, when I clicked in the Quit button, the program just hung. I had to go to the console and and type
quit
to stop the process.
I guess Spyder isn't really intended for GUI programming.

Is there a way to remove the background window that pops up when calling askopenfilename()?

When I call the basic command:
from tkinter.filedialog import askopenfilename
filename = askopenfilename()
A little window pops up behind the selector screen.
Is there a way to remove this?
It pops up because you need at least a "root" window for each tkinter application. In your case, you aren't creating any root window explicitly, so askopenfilename creates it automatically for you. One solution would be to create the root window explicitly and then hide it, something as follows
from tkinter.filedialog import askopenfilename
from tkinter import Tk
Tk().withdraw()
filename = askopenfilename()

Run thread in main window for both Windows and Linux

I am creating a game to run inside a GUI (text area, button, menu etc) I've created a GUI with wxpython. I create a panel inside the main window, which runs a pygame thread.
Problem:
On Windows, the pygame thread runs perfectly inside the main window. But on Linux, the pygame pop up on a new window. How can I set this such that both windows and Linux run the thread in the main window?
Code:
class SDLPanel(wx.Panel):
def __init__(self,parent,ID,tplSize):
global pygame
global pygame_init_flag
wx.Panel.__init__(self, parent, ID, size=tplSize)
self.Fit()
if (sys.platform == 'win32'):
os.environ['SDL_WINDOWID'] = str(self.GetHandle())
os.environ['SDL_VIDEODRIVER'] = 'windib'
else:
os.environ['SDL_VIDEODRIVER'] = 'x11'
#here is where things change if pygame has already been initialized
#we need to do so again
if pygame_init_flag:
#call pygame.init() on subsaquent windows
pygame.init()
else:
#import if this is the first time
import pygame
pygame_init_flag = True #make sure we know that pygame has been imported
pygame.display.init()
window = pygame.display.set_mode(tplSize)
self.thread = SDLThread(window)
self.thread.Start()
def __del__(self):
self.thread.Stop()
print "thread stoped"
#very important line, this makes sure that pygame exits before we
#reinitialize it other wise we get errors
pygame.quit()
Solved problem.
In main window we must self.Show()
Idk why in linux the main window must be showed . Same code.
Tks all
This is a disclaimer alert, according to https://forums.libsdl.org/viewtopic.php?p=39332, the solution works only with SDL 1.2 and not 2.0.

Tkinter GUI doesn't work outside iPython

I'm rather unfamiliar with Python, but am attempting to make a simple gui with tkinter (Python 2.7.x). I have one fully functional button, but I can only get the script to work in iPython. When I try to run it outside of the iPython environment, I see some text run down the command prompt but nothing happens (ie no gui appears, no action happens).
My code:
import Tkinter
from Tkinter import *
import os
#define frame
root = Tk()
frame = Frame(root)
frame.pack()
#define buttons
button = Button(frame, text="Action", command= lambda: os.system("Action.py"))
button.pack(side=LEFT)
Add root.mainloop() to the end of your code.

PyQt4 window not properly closing with signals and slots, only "x" button

As the title suggests, when I create a push button, or a menu option for ending the program, the window doesn't close.
so I'm trying to figure out to end the program and close the window simultaneously. I've been using the tutorial:
http://zetcode.com/tutorials/pyqt4/
which is great otherwise. So how do i connect the push button with ending & closing a widget?
Here is some sample code (copied from the tutorial) that I have been using. I cant seem to get the ending to copy exactly, but I don't think that's the issue anyways:
#!/usr/bin/python
# -*- coding: utf-8 -*-
"""
ZetCode PyQt4 tutorial
This program creates a quit
button. When we press the button,
the application terminates.
author: Jan Bodnar
website: zetcode.com
last edited: October 2011
"""
import sys
from PyQt4 import QtGui, QtCore
class Example(QtGui.QWidget):
def __init__(self):
super(Example, self).__init__()
self.initUI()
def initUI(self):
qbtn = QtGui.QPushButton('Quit', self)
qbtn.clicked.connect(QtCore.QCoreApplication.instance().quit)
qbtn.resize(qbtn.sizeHint())
qbtn.move(50, 50)
self.setGeometry(300, 300, 250, 150)
self.setWindowTitle('Quit button')
self.show()
def main():
app = QtGui.QApplication(sys.argv)
ex = Example()
sys.exit(app.exec_())
if __name__ == '__main__':
main()
Thanks!
The code works. I have the same issue, but running it outside IDLE it works. No need to change the code to call the quit method on the QtGui module.
Your app is a QtGui.QApplication, so why are you connecting to the quit signal using QtCore.QCoreApplication? Changing that to QtGui.QApplication.instance().quit works.
replace qbtn.clicked.connect(QtCore.QCoreApplication.instance().quit) with
qbtn.clicked.connect(self.close)
I too faced the same problem (trying to learn Qt4 from the same zetcode tutorials) and came here to look for a solution

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