tkinter，获取文件目录路径作为pandas的变量

Question

我希望用户从文件资源管理器中选择一个文件，并将其路径获取到变量中，例如：

filepath = "CUserRandom_Fileexcel_archive.xlsx"

并使用该变量将其加载到pandas数据帧中

df = pd.read_excel(filepath)

我现在只有这个：

def fileopen():
filepath = filedialog.askopenfile()
label = Label(text=filepath).pack()
df = pd.read_excel(label)
button_choose_file = Button(window, text='choose file', command = fileopen).pack()

但它不起作用。我该如何解决这个问题？

Answer 1

这可能会奏效。将filedialog.askopenfile()更新为filedialog.askopenfilename()，将pd.read_excel(label)更新为pd.read_excel(filepath)

from tkinter import *
from tkinter import filedialog
import pandas as pd
window = Tk()
def fileopen():
filepath = filedialog.askopenfilename(filetypes=(("xlsx", "*.xlsx"), ("all files", "*.*"))) #===assigns the path to filepath
label = Label(window, text=filepath) #==Adds Label to window 
label.pack()
df = pd.read_excel(filepath) #==reads the excel file by fetching it from its path
button_choose_file = Button(window, text='choose file', command=fileopen)
button_choose_file.pack()
window.mainloop()