演示

Question

为作业设计文本编辑器，并不断获得＆＃39; NameError：全球名称＆＃39;事件＆＃39;没有定义＆＃39; 这个python编程的新手，所以任何帮助都会很棒。我已经删除了一些代码作为脚本很长时间，我认为包含了重要的部分。

from tkinter import *
from tkinter.ttk import *
from tkinter.scrolledtext import ScrolledText
import tkinter.filedialog
from datetime import datetime, date
import tkinter.messagebox as box
import tkinter.font

class Notepad(Frame):
            # initialization. receives the master widget
    def __init__(self, parent):
        Frame.__init__(self, parent)

        self.parent = parent
        self.parent.attributes('-topmost', False)
        self.GUI()


    def GUI(self):
            # Main window and title
        self.parent.title("Assisgnments")
        self.pack(fill=BOTH, expand=True)

        self.columnconfigure(1, weight=1)
        self.columnconfigure(3, pad=7)
        self.rowconfigure(3,weight=1)
        self.rowconfigure(5, pad=7)
            # Menu bar and drop down ADD
        menu = Menu(self.parent)
        self.parent.config(menu=menu)


        edit = Menu(self.parent, tearoff=0)
        edit.add_command(label="Image", command=self.showImg)
        edit.add_command(label="Text", command=lambda:self.showTxt(self))  #
        menu.add_cascade(label="Edit", menu=edit)

        self.popup_menu = Menu(self.parent, tearoff=0)
        self.popup_menu.add_command(label="link")


            # Text area and buttons created, positioned and labeled
        self.label = Label(self, text="Notes", font="Bold")
        self.label.grid(sticky=W, pady=4, padx=5)

        self.text = ScrolledText(self)
        self.text.grid(row=1, column=0, columnspan=2, rowspan=6, padx=5, sticky=W+S+N)



        cbtn = Button(self, text="Save", command=lambda:files.save_text(self))
        cbtn.grid(row=10, column=4, pady=4)

        hbtn = Button(self, text="Open",command=lambda:files.load_file(self), width=10)
        hbtn.grid(row=10, column=0, padx=5)

        self.obtn = Button(self, text="Quit", underline=0, command=self.parent.destroy)
        self.obtn.grid(row=10, column=5, pady=(2,4))

        self.bind("<Button-3>", self.do_popup())

    def do_popup(self):
        self.popup_menu.post(event.x_root, event.y_root)



    def sequence(self,*functions):
        def func(*args, **kwargs):
            return_value = None
            for function in functions:
                return_value = function(*args, **kwargs)
            return return_value
        return func


class files(Notepad):
    def load_file(self):
        self.file_name = tkinter.filedialog.askopenfilename()               #("ProgNotes.txt"),("WebAppNotes.txt")
        infile = open(self.file_name,"r")
        self.text.insert(END, infile.read())
        infile.close()

    def note_text(self, name):
        self.file = open("%sNotes.txt"%name, "r")
        self.text.delete("1.0",END)
        self.text.insert(END, self.file.read())

    def save_text(self):

        self.file_name = tkinter.filedialog.asksaveasfilename()
        outfile = open("%s"%self.file_name, "w")
        outfile.write(self.text.get("1.0", END))
        #outfile = self.text.get(1.0, END)

    def save_notes(self, name):
        with open("%sNotes.txt"%name, "w") as self.file:
            self.file.write(self.text.get("1.0", END))
            self.file.close()

        # New window instance with assignment info



def main():

    root = Tk()
    file_name = None
    root.geometry("725x600+300+300")
    Notepad(root)
    root.mainloop()

if __name__ == '__main__':
    main()

Answer 1

要解决您的问题，您需要更改：

def do_popup(self):
    self.popup_menu.post(event.x_root, event.y_root)

为：

def do_popup(self, event):
    self.popup_menu.post(event.x_root, event.y_root)

和

self.bind("<Button-3>", self.do_popup())

为：

self.bind("<Button-3>", self.do_popup)

演示

这样，您将获得此GUI：

试图在tkinter中获得一个弹出菜单

1 个答案:

演示