Select path to Excel file using TKIinter
Question:
i am trying to select the path to an Excel file using TKinter.
The Idea is to have two browser dialogs opening. One asking to select an Excel file an the other asking to define the destination folder.
import numpy as np
import openpyxl
import matplotlib.pyplot as plt
#import plotly.graph_objects as go
import tkinter
from tkinter import filedialog
import os
root = tkinter.Tk()
root.withdraw()
def open_file():
file = filedialog.askopenfile(mode='r', filetypes=[('Excel Files', '*.xlsx')])
filepath = os.path.abspath(file.name)
def search_for_file_path_dest ():
currdir = os.getcwd()
tempdir = filedialog.askdirectory(parent=root, initialdir=currdir, title='Please select the destination directory')
if len(tempdir) > 0:
print ("You chose: %s" % tempdir)
return tempdir
filepath = open_file()
file_path_variabledest = search_for_file_path_dest()
print(filepath)
print(file_path_variabledest)
The destination folder is selected correctly but the path to the excel file (filepath) is "None".
What am I doing wrong?
Thnak you
Answers:
Function open_file
was nothing returned. You must return filepath
.
def open_file():
file = filedialog.askopenfile(mode='r', filetypes=[('Excel Files', '*.xlsx')])
filepath = os.path.abspath(file.name)
return filepath
i am trying to select the path to an Excel file using TKinter.
The Idea is to have two browser dialogs opening. One asking to select an Excel file an the other asking to define the destination folder.
import numpy as np
import openpyxl
import matplotlib.pyplot as plt
#import plotly.graph_objects as go
import tkinter
from tkinter import filedialog
import os
root = tkinter.Tk()
root.withdraw()
def open_file():
file = filedialog.askopenfile(mode='r', filetypes=[('Excel Files', '*.xlsx')])
filepath = os.path.abspath(file.name)
def search_for_file_path_dest ():
currdir = os.getcwd()
tempdir = filedialog.askdirectory(parent=root, initialdir=currdir, title='Please select the destination directory')
if len(tempdir) > 0:
print ("You chose: %s" % tempdir)
return tempdir
filepath = open_file()
file_path_variabledest = search_for_file_path_dest()
print(filepath)
print(file_path_variabledest)
The destination folder is selected correctly but the path to the excel file (filepath) is "None".
What am I doing wrong?
Thnak you
Function open_file
was nothing returned. You must return filepath
.
def open_file():
file = filedialog.askopenfile(mode='r', filetypes=[('Excel Files', '*.xlsx')])
filepath = os.path.abspath(file.name)
return filepath