tkinter display file name selected as text label in grid-CodePudding

I'm new to Python and I'm trying to display the name of a file selected in a tkinter window using a grid format.

the code I have managed to put together so far is as follows:

from tkinter import *
from tkinter.ttk import *
from tkinter.filedialog import askopenfile 
import time

ws = Tk()
ws.title('Select Input Files')
ws.geometry('400x300') 


def open_file():
    file_path = askopenfile(mode='r', filetypes=[('CSV Files', '*.csv')])
    if file_path is not None:
        pass

up1 = Label(
    ws, 
    text='sales_history_document '
    )
up1.grid(row=0, column=0, padx=10)

btn1 = Button(
    ws, 
    text ='Choose File', 
    command = lambda:open_file()
    )
btn1.grid(row=0, column=1)

up2 = Label(
    ws, 
    text='style_group_document '
    )
up2.grid(row=1, column=0, padx=10)

btn2 = Button(
    ws, 
    text ='Choose File ', 
    command = lambda:open_file()
    ) 
btn2.grid(row=1, column=1)

upld = Button(
    ws, 
    text='Upload Files', 
    command=uploadFiles
    )
upld.grid(row=7, columnspan=3, pady=10)

ws.mainloop()

The output is a series of labels and buttons but I would like to have the file name populate to the left to each button.

view output here

CodePudding user response：

Hey you are almost there. askopenfile gets the filename for you and you can use the name information like in the example below. I used the os package in addition:

from tkinter import *
from tkinter.ttk import *
from tkinter.filedialog import askopenfile 
import time
import os

ws = Tk()
ws.title('Select Input Files')
ws.geometry('400x300') 


def open_file():
    file_path = askopenfile(mode='r', filetypes=[('CSV Files', '*.csv')])

    if file_path is not None:
        # this gets the full path of your selected file
        filename = file_path.name
        # this is only selecting the name with file extension
        filename = os.path.basename(filename)
        # to replace your Label you first have to destroy the other label
        up1.destroy()
        # then create a new label with the filename
        file_label = Label(ws, text=filename).grid(row=0, column=0, padx=10)
        # you can even change the title of your tkinter window if you want
        ws.title('Select Input Files -- '   filename)

up1 = Label(
    ws, 
    text='sales_history_document '
    )
up1.grid(row=0, column=0, padx=10)

btn1 = Button(
    ws, 
    text ='Choose File', 
    command = lambda:open_file()
    )

btn1.grid(row=0, column=1)

up2 = Label(
    ws, 
    text='style_group_document '
    )
up2.grid(row=1, column=0, padx=10)

btn2 = Button(
    ws, 
    text ='Choose File ', 
    command = lambda:open_file()
    ) 
btn2.grid(row=1, column=1)

ws.mainloop()

CodePudding user response：

If you just want to get the filename, use askopenfilename() instead of askopenfile().

You need to pass the label widget to open_file() if you want to show the filename in the label at the left of the button:

...
from tkinter.filedialog import askopenfilename
...
def open_file(lbl):
    file_path = askopenfilename(filetypes=[('CSV Files', '*.csv')])
    if file_path is not None:
        lbl.config(text=file_path)

up1 = Label(
    ws,
    text='sales_history_document '
    )
up1.grid(row=0, column=0, padx=10)

btn1 = Button(
    ws,
    text ='Choose File',
    command = lambda:open_file(up1)
    )
btn1.grid(row=0, column=1)

up2 = Label(
    ws,
    text='style_group_document '
    )
up2.grid(row=1, column=0, padx=10)

btn2 = Button(
    ws,
    text ='Choose File ',
    command = lambda:open_file(up2)
    )
btn2.grid(row=1, column=1)
...