CODEWITHSUNDEEP
pythonpandastkinterttkbootstrap
Richmond Nguyen
Richmond Nguyen

Reputation: 21

Why won't my tkinter code to open a file work?

I am trying to make a application to get a user to press a button and be able to open a csv file and then make a copy of that file in a certain directory, however the first line of code in the function won't work. Here is an excerpt of the code below.

import os
import pandas as pd
from tkinter import filedialog
import ttkbootstrap as tb


root = tb.Window(themename='superhero')

def add_spreadsheet():
    spreadsheet_filepath = filedialog.askopenfilename(initialdir='C:\Users\peter\Downloads',
                                                      title='Select A File', filetypes=
                                                      ('csv files', '*.csv'))
    file_name = os.path.basename(spreadsheet_filepath)
    df = pd.read_csv(spreadsheet_filepath)
    output_path = os.path.join('csv files/', f'{file_name}')
    df.to_csv(output_path, index=False)

# create add button to add a spreadsheet to a file
home_add_btn = tb.Button(root, text='Add', command=add_spreadsheet)
home_add_btn.pack(anchor='center', pady=10)

root.mainloop()

I tried online but I couldn't understand anything.

Upvotes: -1

Views: 108

Answers (3)

Hermann12
Hermann12

Reputation: 3581

There are two things you have to improve:

  1. The backslash in your path will interpreted as an escape sequence. You can prevent this, if you use a raw string. This does a prefixed with a letter r or R in front of the path name.

Other options for path names, but in my opion the raw string is prefered.

  1. I added also brackets around your filetypes, because they are defined as a sequence of tuples, see doc.
spreadsheet_filepath = filedialog.askopenfilename(initialdir= r'C:\Users\peter\Downloads', title='Select A File', defaultextension=".csv", filetypes=[("CSV Files", "*.csv")])

Here is a dummy code what opens the file dialoge:

from tkinter import *
from tkinter import filedialog

root = Tk() 

def add_spreadsheet():
    spreadsheet_filepath = filedialog.askopenfilename(initialdir= r'C:\Users',
                                                      title='Select A File', filetypes=
                                                      [('csv files', '*.csv')])
    file_name = os.path.basename(spreadsheet_filepath)
    df = pd.read_csv(spreadsheet_filepath)
    output_path = os.path.join('csv files/', f'{file_name}')
    df.to_csv(output_path, index=False)

# create add button to add a spreadsheet to a file
home_add_btn = Button(root, text='Add', command=add_spreadsheet)
home_add_btn.pack(anchor='center', pady=10)

root.mainloop()

Upvotes: 1

Luke L
Luke L

Reputation: 313

The problem you are facing is most likely a syntax error, probably like this:

SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in position 2-3: truncated \UXXXXXXXX escape

(At least that's what I get when I run your code). The error message here states that a "unicode" error occurs. The problem with your code is the string literal of the file path in the first line of your add_spreadsheet function. Specifically, this line:

spreadsheet_filepath = filedialog.askopenfilename(initialdir='C:\Users\peter\Downloads', title='Select A File', filetypes=('csv files', '*.csv'))

Normal strings will interpret backslashes (\) and subsequent character(s) as something called "escape sequences". For example, the string "Hello World\nHello World again" will be interpreted as "Hello World" then a newline character, and then finally "Hello World again". Some common escape sequences include \n (newline), \t (tab), \r (line feed), and \uxxx (unicode character with ordinal number xxx). If the character after the backslash is not one of the valid escape sequences, then Python will raise a unicode error. Unfortunately, file paths are usually given with backslash characters inside. To solve this issue, there are two approaches:

  1. Use double backslashes (\\). This will escape the escape sequence and make it valid. Example:
spreadsheet_filepath = filedialog.askopenfilename(initialdir='C:\\Users\\peter\\Downloads', title='Select A File', filetypes=('csv files', '*.csv'))

This is fine from a syntactical viewpoint, and may be the only choice if other backslashes are meant for actual escape sequences, like the newline (\n). But if not, the second approach works better.

  1. Use raw strings. They are called "raw" strings because they do not convert backslashes and escape sequences into their actual form. Thus, any number of backslashes and escape sequences that are otherwise invalid will not raise an error. Change your code to this:
spreadsheet_filepath = filedialog.askopenfilename(initialdir=r'C:\Users\peter\Downloads', title='Select A File', filetypes=('csv files', '*.csv'))

This approach not only saves a lot of typing if there are many invalid escapes but also makes clear that this string is "raw". Thus, I highly recommend the latter approach. However, if you want a newline or other valid escapes then you need to use the first method.

Upvotes: 1

Lagor845
Lagor845

Reputation: 31

In strings the "" character has special properties to it. To fix your current issue just use "\\" instead...

For example:

spreadsheet_filepath = filedialog.askopenfilename(initialdir='C:\\Users\\peter\\Downloads',
                                                      title='Select A File', filetypes=
                                                      ('csv files', '*.csv'))

Just replace your code with the code segment above and it should work.

Upvotes: 1

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