I want tkinter button to work like a messagebox.

Asked 2 years ago, Updated 2 years ago, 103 views

I'm a python beginner.
Using Button on tkinter, I am trying to create a program where the window disappears after storing different variables (True, False) depending on the button I click (OK, cancel).The movement is like messagebox's askokcancel.
Normally, I would use messagebox, but I wanted to include hyperlinks in the dialog, so I made it with Button as follows.
This is the main topic, but with this source, the return buff will be returned as True no matter which button you press.It seems that it will not be possible to store it in BooleanVar after destroyed.
If you turn off destroy, the return value will change depending on the button, but of course the frame will not disappear.
Could you please let me know if there is a good way to make them coexist?

Python 3.5.2::Anaconda 4.0.0 (32-bit)
Tcl/Tk 8.6
That's it.

import tkinter ask
import webbrowser

def hyperlink (event):
    webbrowser.open_new(event.widget.cget('text'))

def program_start():
    root.destroy()
    return lambda:buff.set(True)

def program_quit():
    root.destroy()
    return lambda:buff.set(False)

root=tk.Tk()
buff=tk.BooleanVar()
buff.set (True)

lbl=tk.Label (root, text=r'https://google.co.jp', fg="blue", cursor="hand2")
lbl.pack()
lbl.bind("<Button-1>", hyperlink)
tk.Button(root, text='OK', width=10, command=program_start).pack()
tk.Button(root, text='cancel', width=10, command=program_quit).pack()

root.mainloop()

print(buff.get())

python3 tkinter

2022-09-30 17:26

2 Answers

As far as the reference, callback doesn't seem like a lambda formula in the first place...

As a solution, isn't it possible to do destroy after set?

def program_start():
    buff.set (True)
    root.destroy()

def program_quit():
    buff.set (False)
    root.destroy()

Then

If you turn off destroy, the return value will change depending on the button.

Assuming that erasing destroy is equivalent to commenting out root.destroy(), wouldn't the code that is set simply return as part of the lambda formula?I don't actually reproduce it on hand.


2022-09-30 17:26

If it is classified and reused, I think I can write it like this.

import tkinter ask
import webbrowser

classMsgBox():
    def__init__(self):
        self.root=tk.Tk()
        self.ret = False
        lbl=tk.Label(self.root, text=r'https://google.co.jp',fg="blue", cursor="hand2")
        lbl.pack()
        lbl.bind("<Button-1>",self.hyperlink)
        tk.Button(self.root,text='OK',width=10,command=lambda:self.root_exit(True))).pack()
        tk.Button(self.root,text='cancel',width=10,command=lambda:self.root_exit(False))).pack()

    def hyperlink (self, event):
        webbrowser.open_new(event.widget.cget('text'))

    def start (self):
        self.root.mainloop()

    def root_exit(self, state):
        self.ret=state
        self.root.destroy()

    default(self):
        return self.ret

msg_box=MsgBox()
msg_box.start()
print(msg_box.ret)


2022-09-30 17:26

If you have any answers or tips


© 2024 OneMinuteCode. All rights reserved.