How can I change a text label from another file in python tkinter?-CodePudding

I hope you can help me, because I am stuck on this. I have an application on tkinter, bassically I need that when pressing a button a function from another python file is executed and changes the text of a label.

Let me explain:

My frontend file is called ODF_principal.py and my backend file is called ODF_second.py, in the backend file I have functions.

I imported ODF_second.py asOFU on ODF_principal to set command in button properties like the following:

btn_Open = Button(frame_menu, text="Open file", command=OFU.open_file)

btn_Open belongs to frontend

Well, when I run the function, I need that when a file is opened or not, change the text of a label that belongs to the frontend file.

lbl_file = Label(frame_menu)

I've tried to import the frontend file into de backend but I've got an error of CIRCULAR IMPORT, so here is the function that I need it to be able to change the text of lbl_file

opened_file_path=""

def open_file():
global opened_file_path
opened_file_path=filedialog.askopenfilename(initialdir = "/",
            title = "Select files",filetypes = (("TXT files","*.txt"),
            ("All files","*.*")))
if not opened_file_path:
    opened_file_path = ""
**OFR.lbl_file.config(text=opened_file_path)**

Thank you for your time.

CodePudding user response：

You can achieve this using a lambda function:

btn_Open = Button(frame_menu, text="Open file", command= lambda: OFU.open_file(lbl_file))

And then

def open_file(label):
    global opened_file_path
    opened_file_path=filedialog.askopenfilename(initialdir = "/",
    title = "Select files",filetypes = (("TXT files","*.txt"),
            ("All files","*.*")))
    if not opened_file_path:
        opened_file_path = ""
    label.config(text=opened_file_path)

The lambda function allows you to pass a reference to the label to your function in the other file.

CodePudding user response：

You can achieve this by getting open_file to return the file path and altering the label in your frontend.

def button_press():
    file_open = OFU.open_file
    my_label.configure(text=file_open)

btn_Open = Button(frame_menu, text="Open file", command=button_press)

And

def open_file():
    opened_file_path=filedialog.askopenfilename(initialdir = "/",
                title = "Select files",filetypes = (("TXT files","*.txt"),
                ("All files","*.*")))
    if not opened_file_path:
        opened_file_path = ""
    return opened_file_path

This is an alternative to @Henry's answer that means your backend doesn't have to handle any of the GUI