【发布时间】:2017-06-10 00:51:22
【问题描述】:
NameError: 名称“onOpen”未定义
命令功能有问题。我不确定我在这里做错了什么。我在 onOpen 函数之前测试了代码,它工作正常。
import tkinter as tk
from tkinter import filedialog
class Application(tk.Frame):
def __init__(self, master=None):
tk.Frame.__init__(self, master)
self.pack()
self.createWidgets()
def onOpen():
""" Ask the user to choose a file and update the values of label change button"""
button_label.set(filedialog.askopenfilename(filetypes = ()))
def createWidgets(self):
#instruction label
self.labelInstruct = tk.Label(self, text="Instructions:", padx=10, pady=10)
self.labelInstruct = tk.Label(self, text="All you have to do is insert the file and save it.\n The conversion is instant", padx=10, pady=10)
self.labelInstruct.pack()
#insertfile button
self.ifbut = tk.Button(self, text="Insert File", command=onOpen)
self.ifbut.pack()
button_label = tk.StringVar(self)
text = tk.Label(self, textvariable = button_label)
text.pack()
#save button
self.saveLabel = tk.Label(self, text="Save File", padx=10, pady=10)
self.saveLabel.pack()
self.saveButton = tk.Button(self)
self.saveButton.pack()
#quit button
self.quitButton = tk.Button(self, text='Quit',
command=self.quit)
self.quitButton.pack()
app = Application()
app.master.title('Sample application')
app.mainloop()
【问题讨论】:
标签: python python-2.7 python-3.x tkinter