Develop Reference

How to use a python application (tkinter).exe to open a text file

im a beginner at python and im trying to created a program using tkinter that view and edit text. i used pyinstaller to make an .exe file to view text, but when i used "open with" to a text file and select my text viewer program nothing shows.
then i noticed that how is my program supposed to open the file and display its contents. because originally i have this button "open file" that when clicked it, it will ask me what to open, but now that opening the file comes first before the program runs, i cannot click the button so my program does not know what to do. is there any way for my program to know what am i trying to open? do i have to import something?
thanks for the answer
from tkinter import *
from tkinter import filedialog
root=Tk()
def openfile():
textname=filedialog.askopenfilename(title="Open File", filetypes=(("Text", "*.txt"), ("Python", "*.py"), ("Html", "*.html"), ("All Files", "*.*")))
openedfile=open(textname,'r')
content=openedfile.read()
n_text.insert(END,content)
openedfile.close()
n_text=Text(root, font=11, relief=FLAT)
n_text.pack()
btn=Button(root, text="open", command=openfile).pack()
root.mainloop()

Not sure if I understood your question well. Do you need to know the path of the file you are opening?
from os.path import split
textname=filedialog.askopenfilename(title="Open File", filetypes=(("Text", "*.txt"), ("Python", "*.py"), ("Html", "*.html"), ("All Files", "*.*")))
pathname = split(textname)
#This print is just to show your path and filename
print("path:", pathname[0], "file:",pathname[1])
Based on your comments I add a new answer it take the full path from arguments on scripts.
from tkinter import *
from tkinter import filedialog
import sys
root=Tk()
def openfile():
textname=filedialog.askopenfilename(title="Open File", filetypes=(("Text", "*.txt"), ("Python", "*.py"), ("Html", "*.html"), ("All Files", "*.*")))
openedfile=open(textname,'r')
content=openedfile.read()
n_text.insert(END,content)
openedfile.close()
n_text=Text(root, font=11, relief=FLAT)
n_text.pack()
btn=Button(root, text="open", command=openfile).pack()
if len(sys.argv) == 2:
openedfile=open(sys.argv[1],'r')
content=openedfile.read()
n_text.insert(END,content)
openedfile.close()
root.mainloop()
I added the path in this format:
c:/Users/user/Python/pruebas.py "C:\Users\user\Desktop\text_file.txt"

Opening a text file with your python script

I have created a program that can edit the contents of a text file. Assume that my program looks something like this:
from tkinter import *
from tkinter import filedialog
root = Tk()
def open_file():
file_to_open = filedialog.askopenfilename(initialdir="C:/" , filetypes=(("All files" , "*.*"),))
if file_to_open != "":
file = open(file_to_open , 'r')
content_in_file = file.read()
file.close()
text.delete('1.0' , END)
text.insert('1.0' , content_in_file)
def save_file():
path = filedialog.asksaveasfilename(initialdir="C:/" , filetypes=(("All files" , ""),))
if path != "":
file = open(path , 'w')
file.write(text.get('1.0' , 'end-1c'))
file.close()
text = Text(root , width = 65 , height = 20 , font = "consolas 14")
text.pack()
open_button = Button(root , text = "Open" , command = open_file)
open_button.pack()
save_button = Button(root , text = "Save" , command = save_file)
save_button.pack(pady=20)
mainloop()
The problem here is that when I click on a text file in my file explorer, it opens with the default windows notepad instead of opening with my program.
What I want is that all the text files should open with my program instead of opening with the default windows Notepad.
Here's what I did (in order):
After completing the following steps, I tried opening my text file, but it says:
I tried converting my python program into an exe file (using pyinstaller) and then following the steps above, but when I open the text file I get an other error:
Is there anything wrong with my code or the steps I followed?
I would really appreciate it if anyone could guide me through how I can open a text file with my program.

The code looks fine, it just needed to take the argument, when you open-with you are calling some executable on a path o multiple paths, the first argument is de executable itself, so if the code is executed with more than 1 argument it's a path; the problem with this is that the command is python and not file.py, a fix would be convert it to a exe or call it with a bat.
This example may look different but is more or less the same, just packed inside a class.
from tkinter import filedialog
import tkinter as tk
import sys
import os
SRC_PATH = os.path.dirname(__file__)
class File_Editor(tk.Tk):
def __init__(self):
tk.Tk.__init__(self)
self.text = tk.Text(self, width=65, height=20, font="consolas 14")
self.text.pack()
self.open_button = tk.Button(self, text="Open", command=self.open_file)
self.open_button.pack()
self.save_button = tk.Button(self, text="Save", command=self.save_file)
self.save_button.pack(pady=20)
def open_file(self, file=None):
path = file or filedialog.askopenfilename(
initialdir=SRC_PATH, filetypes=(("All files", "*.*"),))
if os.path.exists(path) and os.path.isfile(path):
with open(path, 'r', encoding='utf-8') as file:
content_in_file = file.read()
self.text.delete('1.0', tk.END)
self.text.insert('1.0', content_in_file)
def save_file(self):
path = filedialog.asksaveasfilename(initialdir=SRC_PATH, filetypes=(("All files", ""),))
if os.path.exists(path) and os.path.isfile(path):
with open(path, 'r', encoding='utf-8') as file:
file.write(self.text.get('1.0', 'end-1c'))
if __name__ == '__main__':
app = File_Editor()
if len(sys.argv) > 1:
app.open_file(sys.argv[1])
app.mainloop()
This .bat file is just passing the firs argument to the file; a full path is required to execute, it will not be called in a directory you expect.
#echo off
python D:\Escritorio\tmp_py\temp_3.py %1
Now you just call the open-with File_Editor.bat

How to use filedialog to open an excel file and pandas to save/read/write that file

I am very new to Python so apologizing up front for a stupid question. I am building a small application that allows me to first seach for an excel file (with filedialog) from my computer, imports it into the program and then I could do something with it.
So the application has a button "search file" and "save file", but the save_file function doesnt work. It works if I put it inside the open_file function.
Am I going completely to wrong way here? I tried to google like crazy, but didnt find an answer for this
Thanks a lot for your support!
-KV
""" IMPORT AND SAVE XLSX FILE """
def open_file():
filename = filedialog.askopenfilename(title="Select file", filetypes=(("Excel files", "*.xlsx"),("Excel files", ".xls")))
file_entry.insert(END, filename)
def save_file():
file_data = pd.read_excel(filename)
print(file_data)
""" UI SETUP """
file_entry = Entry(width=60)
file_entry.grid(row=1, column=1)
file_button = Button(text="Select File", width=13, command=open_file)
file_button.grid(row=1, column=2)
file_button = Button(text="Save File", width=13, command=save_file)
file_button.grid(row=1, column=3)

In save_file() the variable filename isn't defined. You have to call the function with an argument e. g. save_file(filename) or define it inside the function.

How to open file with tkfiledialog and read the content with notepad

I want to read the content of the selected file by opening with tk filedialog. When i select the file and click on open button the file doesn't open but rather close the dialog box .How can i open the selected file with notepad so i can be able to read the content in the file.
from tkinter import *
from tkinter import filedialog
def my_file():
filename = filedialog.askopenfile(mode="r", initialdir="/", title="select file",
filetypes=(("text files", "*.txt"), ("all files", "*.*")))
root = Tk()
root.geometry("300x300")
#open the selected txt file with notepad to read the content
b = Button(root, text="open text file", command = my_file).pack()
root.mainloop()
EDIT
With hint from #PM 2RING and #Erik i changed the filedialog.askopenfile to filedialog.askopenfilename to return it to open with notepad.exe when i select the file.
THis is the code:
from tkinter import *
from tkinter import filedialog
import os
def my_file():
filename = filedialog.askopenfilename( initialdir="C:/", title="select
file", filetypes=(("text files", "*.txt"), ("all files", "*.*")))
for f in filename:
return f
os.system(r"C:/notepad.exe" + f)
root = Tk()
root.geometry("300x300")
#open the selected txt file with notepad to read
the content
b = Button(root, text="open text file", command = my_file).pack()
root.mainloop()
it output this error :
Blockquote'C:/notepad.exet' is not recognized as an internal or external command,
operable program or batch file.
Blockquote
but when i changed the return to print it print the directory to terminal.I tried to open with subprocess
subprocess.Popen([r'C:\Program Files (x86)\Notepad.exe' + f])
it also doesn't open with this one to.

There's a few things which need to be amended here.
First of all, C:/notepad.exe is not the location of Notepad (At least not on any Windows machine with a default setup), you can simply use notepad.exe which should make it more compatible with systems that have moved Notepad to another location (citation needed).
Secondly, executing . . .
for f in filename:
return f
os.system(r"C:/notepad.exe" + f)
Doesn't do what you seem to think it does. What's actually happening here is that your program is loading the string into the loop, evaluating the first character (Probably "C") and then returning the string to the Button widget which doesn't receive any returned values. This then breaks your function, so it never actually reaches your declaration of os.system(r"C:/notepad.exe" + f).
You also need to include a space between the statement used to open Notepad notepad.exe and the actual file declaration f, otherwise you're running something like notepad.exeC:/text.txt which is going to throw an error at you.
What you should be doing is something like the below:
from tkinter import *
from tkinter import filedialog
import os
def my_file():
filename = filedialog.askopenfilename( initialdir="C:/", title="select file", filetypes=(("text files", "*.txt"), ("all files", "*.*")))
os.system(r"notepad.exe " + filename)
root = Tk()
root.geometry("300x300")
b = Button(root, text="open text file", command = my_file).pack()
root.mainloop()
I'd like to add that I have no clue why you're doing this, why not simply display the text in a Text widget? What you're doing here is adding one extra step for people to open files in Notepad, instead of opening file explorer, finding the file and then opening it they have to open your program, then open file explorer, then find the file and then open it. It's adding at least one extra click not to mention the load time of your program.

As mentioned by PM 2Ring, I would use the os.system function. As mentioned in its description, "os.system(command)" let's you execute the command as if you had written it in the command prompt, so os.system("Notepad c:/users/yourName/junk.txt) would open a file named junk.txt if it were at that location.
In other words, once you have the filename from your filedialog.askopenfilename call, do something like this:
import os
os.system("Notepad " + filename) # where filename is the string that filedialog.askopenfilename returns
The implementation into your code shouldn't be too bad, but if you need a more complete example, let let me know.

The following code will display a button and a text widget. The button will load in your file, and the text widget will display it. I assume this is what you had in mind?
from tkinter import *
from tkinter import filedialog
def my_file():
file = filedialog.askopenfile(mode="r", initialdir="/", title="select file",
filetypes=(("text files", "*.txt"), ("all files", "*.*")))
t.insert(END, file.read())
file.close()
root = Tk()
root.geometry("300x300")
#open the selected txt file with notepad to read the content
t = Text(root)
t.grid(row=0, column=0)
b = Button(root, text="open text file", command = my_file)
b.grid(row=1, column=0)
root.mainloop()

I want to use extensions list at the time of "save as" option in File Dialog for file write in Python 3

I want to use extensions list example : [".txt",".html",".css"] for save as option in file dialog popup window. when I use this method
file_opt = options = {}
options['defaultextension'] = '.txt',I can able to write any file with .txt as default extension without choose in save as option but I want to choose extensions for save as in file dialog popup window from using my extensions list.
I'm using Python 3.5 based Anaconda IDE

If you look at the documentation here you can see that you can pass in the filetypes keyword which specifies a list of tuples that have the name, and file extension respectively for the different types of filetypes you want to be able to save as.. So you can do something along the lines of:
import tkinter as tk
from tkinter import filedialog as fd
def save_file():
filename = fd.asksaveasfilename(defaultextension='.txt',
filetypes= [('Text','.txt'), ('HTML', '.html'), ('CSS', '.css')])
if filename:
print("User saved the filename with extension:", filename.split(".")[-1])
root = tk.Tk()
button = tk.Button(root, text='Save File', command=save_file)
button.pack()
root.mainloop()
Or if you really want to use a dictionary for this:
import tkinter as tk
from tkinter import filedialog as fd
SAVE_OPTS = {'defaultextension':'.txt',
'filetypes': [('Text','.txt'), ('HTML', '.html'), ('CSS', '.css')]}
def save_file():
filename = fd.asksaveasfilename(**SAVE_OPTS)
if filename:
print("User saved the filename with extension:", filename.split(".")[-1])
root = tk.Tk()
button = tk.Button(root, text='Save File', command=save_file)
button.pack()
root.mainloop()

SaveFileDialog sd = new SaveFileDialog();
sd.Filter = "Text File (*.txt)|*.txt|PNG File (*.txt)|*.png"|...;
if(sd.ShowDialog() == DialogResult.OK) {
richTextBox1.SaveFile(sd.FileName, RichTextBoxStreamType.PlainText);
}