[已解决]打开文件报错

召唤狮

from tkinter import *
root = Tk()
def callback():
     fileName = filedialog.askopenfilename()
     print(fileName)
Button(root, text="打开文件", command=callback).pack()
mainloop() #点打开文件按钮报错

报错内容：
Exception in Tkinter callback
Traceback (most recent call last):
  File "D:\anaconda3\lib\tkinter\__init__.py", line 1883, in __call__
    return self.func(*args)
  File "<pyshell#5>", line 2, in callback
NameError: name 'filedialog' is not defined

最佳答案

月排行榜 / 总排行榜

Twilight6

2021-2-5 14:49:37

filedialog 等一些组件在 * 导入是不被导入的，要重新进行导入

参考代码：

from tkinter import *
from tkinter import filedialog

root = Tk()
def callback():
     fileName = filedialog.askopenfilename()
     print(fileName)
Button(root, text="打开文件", command=callback).pack()
mainloop() #点打开文件按钮报错

跳转到最佳答案楼层

Daniel_Zhang

from tkinter import filedialog

Twilight6

filedialog 等一些组件在 * 导入是不被导入的，要重新进行导入

参考代码：

from tkinter import *
from tkinter import filedialog

root = Tk()
def callback():
     fileName = filedialog.askopenfilename()
     print(fileName)
Button(root, text="打开文件", command=callback).pack()
mainloop() #点打开文件按钮报错

逃兵

from tkinter import *
from tkinter import filedialog

root = Tk()
def callback():
     fileName = filedialog.askopenfilename()
     print(fileName)
Button(root, text="打开文件", command=callback).pack()
mainloop() #点打开文件按钮报错