如果在 Python 中选择了下拉列表中的项目，则提示用户输入

Question

I have a GUI in python that, amongst other things, let's the user choose items from a dropdown menu (I used the combobox feature from tkinter).

I wish to have it so that when the item "Custom" is select, an input box appears to ask the user what their custom number would be. I don't want to have to use a button for the box to appear, but somehow have it so that as soon as custom is selected, the input box appears. I tried first with while loops but I can't get it to work, same with if statements :s

I also tried using the askinteger() method that I found here ( http://effbot.org/tkinterbook/tkinter-entry-dialogs.htm ) but none of the combinations I came up with worked (I am very new at Python and am still learning so excuse my obvious mistakes).

Here is my code for the GUI :

from tkinter import *
from tkinter.ttk import *
from tkinter import filedialog
from tkinter import StringVar
from tkinter import messagebox
from tkinter import simpledialog


class MyGUI:
    def __init__(self, master):

        self.master = master
        master.title("Intent/Interpretation Check")

        self.runlabel = Label(master, text="RunID :")
        self.runlabel.grid(row=0, column=0)

        self.runentry = Entry(master)
        self.runentry.grid(row=1, column=0, padx=25)

        self.checklabel = Label(master, text="Check type :")
        self.checklabel.grid(row=0, column=1)

        self.typeselect = Combobox(master)
        self.typeselect['values']=("Intent Score", "Interpretation Score")      
        self.typeselect.grid(row=1, column=1, padx=25)

        self.limitlabel = Label(master, text="Fails if score is below :")
        self.limitlabel.grid(row=0, column=2, padx=25)

        self.limitselect = Combobox(master)
        self.limitselect['values']=(1000, 5000, "Custom")       
        self.limitselect.grid(row=1, column=2, padx=25)
        if self.limitselect.get() != "Custom":
            self.limit = self.limitselect.get()
            pass
        else:
            self.askinteger("Custom limit", "Please enter a number from 1 to 10000", minvalue=1, maxvalue=10000)

        self.submitbutton = Button(master, text="Submit", command=self.checkstatus)
        self.submitbutton.grid(row=1, column=3, padx=25, pady=5)

root = Tk()
root.geometry("+600+300")
my_gui = MyGUI(root)
root.mainloop()

Thank you very much in advance !

Answer 1

You need to have a Bool that tells when to show the new input should be shown. You also need to be constantly polling the ComboBox to see if it's value is equal to "Custom". This is what I came up with in about 3 minutes.

I didn't try to make the GUI look pretty, just a functional example.

from tkinter import *
from tkinter.ttk import *

class Gui:

    def __init__(self):
        self.root = Tk()

        # Set up the Combobox
        self.selections = Combobox(self.root)
        self.selections['values'] = ['Apples', 'Oranges', 'Blueberries', 'Bananas', 'Custom']
        self.selections.pack()

        # The Entry to be shown if "Custom" is selected
        self.custom_field = Entry(self.root)
        self.show_custom_field = False

        # Check the selection in 100 ms
        self.root.after(100, self.check_for_selection)

    def check_for_selection(self):
        '''Checks if the value of the Combobox equals "Custom".'''


        # Get the value of the Combobox
        value = self.selections.get()

        # If the value is equal to "Custom" and show_field is set to False
        if value == 'Custom' and not self.show_custom_field:

            # Set show_field to True and pack() the custom entry field
            self.show_custom_field = True
            self.custom_field.pack()


        # If the value DOESNT equal "Custom"
        elif value != 'Custom':

            # Set show_field to False
            self.show_custom_field = False

            # Destroy the custom input
            self.custom_field.destroy()

            # Set up a new Entry object to pack() if we need it later.
            # Without this line, tkinter was raising an error for me.
            # This fixed it, but I don't promise that this is the
            # most efficient method to do this.
            self.custom_field = Entry(self.root)

        # If the value IS "Custom" and we're showing the custom_feild
        elif value == 'Custom' and self.show_custom_field:
            pass


        # Call this method again to keep checking the selection box
        self.root.after(100, self.check_for_selection)


app = Gui()
app.root.mainloop()

Hope this helps!

EDIT:

To open a new window instead of packing it inside the same window as the Combobox, replace the function check_for_selection with this:

def check_for_selection(self):
    value = self.selections.get()

    # If the value is equal to "Custom" and show_field is set to False
    if value == 'Custom' and not self.show_custom_field:

        # Set show_field to True and pack() the custom entry field
        self.show_custom_field = True

        # Create a new window how we did when we made self.root
        self.new_window = Tk()

        # Create the Entry that will go in the window. The previous Entry widget from line 16, can be removed
        self.custom_field = Entry(self.new_window)
        self.custom_field.pack()

        # Run the new window like we did the original
        self.new_window.mainloop()


    # If the value DOESNT equal "Custom"
    elif value != 'Custom':

        # Destroy the new window that was created if it exists
        if self.show_custom_field:
            self.new_window.destroy()

        # Set show_field to False
        self.show_custom_field = False

    # If the value IS "Custom" and we're showing the custom_feild
    elif value == 'Custom' and self.show_custom_field:
        print('yes')


    # Call this method again to keep checking the selection box
    self.root.after(100, self.check_for_selection)