As .exe , get_current_directory problem?

Python Help
1

Thanks for your Help…

I having a problem with “get_current_directory” in here: INDEX-it-4.0**.exe** .
As .exe it shows current_directory as: C:\Users\vmars\AppData\Local\Temp_MEI224762

When I run it as INDEX-it-4.0**.py** , all is well :
current_directory shows as : ‘C:\Index-it-py’ , as it should .

Pls , how to fix ?

in this code:

    def start_program(self):
        self.file_content = []
        filename = filedialog.askopenfilename(initialdir=self.get_current_directory(), title="Select a file to open")
        if filename:
            with open(filename, 'r') as file:
                self.input_text.delete('1.0', tk.END)
                self.middle_listbox.delete(0, tk.END)  # Clear the listbox before populating
                line_number = 1
                added_words = set()
                for line in file:
                    line_content = line.strip()
                    self.file_content.append(line_content)
                    self.input_text.insert(tk.END, f"{line_number} {line_content}\n")
                    # Populate the middle_listbox with unique words
                    words = line_content.split()
                    for word in words:
                        if word not in added_words:
                            self.middle_listbox.insert(tk.END, word)
                            added_words.add(word)
                    line_number += 1
            self.update_line_count()

I am wondering would I be better off using QT instead of TK .

2

And what exactly is this get_current_directory() method?

3

The correct method should be in the docs for whatever tool has packaged the Python code into an .exe.

If pathlib.Path.cwd() doesn’t work in this non-standard environment, if a Python run-time is bundled in the .exe, maybe sys.executable is a better reference than __file__, (to put into pathlib.Path(...).parent). Or you could try subprocess.run(‘cd’) (or ‘pwd’ on Posix).

4

pathlib.Path.cwd() will always work in that it will always obtain the current working directory. The problem is that the current working directory (i.e. wherever you opened your terminal, or wherever the executable is located if launched by the desktop on Windows, or usually the home directory on other platforms) is rarely what anyone wants.

The MEI directory isn’t ever the default working directory so I doubt this current_directory() features cwd().

1 Like
5

I got things working by code below:

INSTALL-os.cwd-TK-PATH-01.bat

echo
set PATH=%PATH%;C:\Users\vmars\AppData\Local\Packages\PythonSoftwareFoundation.Python.3.12_qbz5n2kfra8p0\LocalCache\Local-packages\Python312\Scripts
pyinstaller --onefile --windowed C:\Index-it-py\~~~INDEX-it-TK-MINI.py
pause

INDEX-it-TK-MINI.py

import tkinter as tk
from tkinter import filedialog, messagebox
import os
class IndexItApp:
    def __init__(self, master):
        self.master = master
        master.title("Index-It-TK-MINI")
        master.resizable(True, True)

        # Address entry for displaying the current working directory
        self.address_entry = tk.Entry(master, width=70)
        self.address_entry.insert(0, self.get_current_directory())
        self.address_entry.grid(row=0, column=1, padx=10, pady=10, sticky="ew")

        # Start button
        self.start_button = tk.Button(master, text="START", command=self.start_program)
        self.start_button.grid(row=1, column=0, padx=10, pady=10)

    def get_current_directory(self):
        try:
            return os.getcwd()  # Default to current working directory
        except Exception as e:
            messagebox.showerror("Error", f"Failed to get the current directory: {e}")
            return ""

    def start_program(self):
        # Opens a file dialog to select a file
        filename = filedialog.askopenfilename(initialdir=self.get_current_directory(), title="Select a file to open")

        if filename:
            try:
                with open(filename, 'r') as file:
                    content = file.read()
                    messagebox.showinfo("File Content", f"File '{os.path.basename(filename)}' opened successfully!")
            except Exception as e:
                messagebox.showerror("Error", f"Failed to open file: {e}")


if __name__ == "__main__":
    root = tk.Tk()
    app = IndexItApp(root)
    root.mainloop()
1 Like