Kivy Python: Detect backspace in Text Input

Question

I am trying to create a simple TextInput for dates that limits input to numbers and auto-populates the the forward slashes for a (mm/dd/yy) format. I was successful in creating a filter that does this by redefining insert_text(), except when the user backspaces, I want to automatically remove the slashes as well. But I don't know how to detect when the user backspaces in the text input so that I can fire an event to erase the slashes if necessary.

Here's a snippet that explains what I want to do, but TextInput does not have an "on_key_up" attribute. Is there a way to add one? Or a better way to go about this?

# .kv file

<DateInput>
    on_key_up: self.check_for_backspace(keycode) # not a true attribute

# .py file

class DateInput(TextInput):

    # checks if last character is a slash and removes it after backspace keystroke.  Not sure this would work.
    def check_for_backspace(self, keycode):
        if keycode[1] == 'backspace' and self.text[-1:] == '/':
            self.text = self.text[:-1]

    #filter for date formatting which works well aside from backspacing
    pat = re.compile('[^0-9]')
    def insert_text(self, substring, from_undo=False):
        pat = self.pat
        if len(substring) > 1:
            substring = re.sub(pat, '', (self.text + substring))
            self.text = ''
            slen = len(substring)
            if slen == 2:
                s = substring[:2] + '/'
            elif slen == 3:
                s = substring[:2] + '/' + substring[2:]
            elif slen == 4:
                s = substring[:2] + '/' + substring[2:] + '/'
            else:
                s = substring[:2] + '/' + substring[2:4] + '/' + substring[4:8]
        elif len(self.text) > 9:
            s = ''
        elif len(self.text) == 1:
            s = re.sub(pat, '', substring)
            if s != '':
                s = s + '/'
        elif len(self.text) == 4:
            s = re.sub(pat, '', substring)
            if s != '':
                s = s + '/'
        else:
            s = re.sub(pat, '', substring)
        return super(DateInput, self).insert_text(s, from_undo=from_undo)

eyllanesc · Accepted Answer · 2018-07-01T06:11:32.923

Since version 1.9.0 TextInput is inherent to FocusBehavior so if you want to detect when you press backspace you must use the keyboard_on_key_down() method or keyboard_on_key_up() method:

from kivy.app import App
from kivy.uix.textinput import TextInput

class DateInput(TextInput):
    def keyboard_on_key_down(self, window, keycode, text, modifiers):
        if keycode[1] == "backspace":
            print("print backspace down", keycode)
        TextInput.keyboard_on_key_down(self, window, keycode, text, modifiers)

    def keyboard_on_key_up(self, window, keycode, text, modifiers):
        if keycode[1] == "backspace":
            print("print backspace up", keycode)
        TextInput.keyboard_on_key_down(self, window, keycode, text, modifiers)


class MyApp(App):
    def build(self):
        return DateInput()

if __name__ == '__main__':
    MyApp().run()

Awesome! This is super helpful. I will switch that out with keyboard_on_key_up(self, window, keycode) in order to prevent running code before the user is finished backspacing. Thank you! — brother_friend_, Jun 30 '18 at 22:05

score 0 · Answer 2 · answered Jun 30 '18 at 22:38

Here's the full solution for anyone wanting a auto-formatted Date Input:

class DateInput(TextInput):
    def keyboard_on_key_up(self, window, keycode):
        if keycode[1] == "backspace" and len(self.text) >= 1:
            if self.text[-1] == "/":
                self.text = self.text[:-1]
            else:
                pass
        else:
            pass
        TextInput.keyboard_on_key_up(self, window, keycode)

    pat = re.compile('[^0-9]')
    def insert_text(self, substring, from_undo=False):
        pat = self.pat
        if len(substring) > 1:
            substring = re.sub(pat, '', (self.text + substring))
            self.text = ''
            slen = len(substring)
            if slen == 2:
                s = substring[:2] + '/'
            elif slen == 3:
                s = substring[:2] + '/' + substring[2:]
            elif slen == 4:
                s = substring[:2] + '/' + substring[2:] + '/'
            else:
                s = substring[:2] + '/' + substring[2:4] + '/' + substring[4:8]
        elif len(self.text) > 9:
            s = ''
        elif len(self.text) == 2:
            s = re.sub(pat, '', substring)
            if s != '':
                s = '/' + s
        elif len(self.text) == 5:
            s = re.sub(pat, '', substring)
            if s != '':
                s = '/' + s
        else:
            s = re.sub(pat, '', substring)
        return super(DateInput, self).insert_text(s, from_undo=from_undo)

score 0 · Answer 3 · answered Jul 10 '18 at 14:58

Solution

Override the do_backspace() method in TextInput. If the text is '/' then return True, indicating that we have consumed the backspace and don’t want it to propagate any further.

Finally, if the text is not '/', we call the original event using super(…) and return the result. This allows the do_backspace event propagation to continue as it would normally have occurred.

Please refer to example for details.

Text Input » do_backspace() method

do_backspace(from_undo=False, mode='bkspc')
Do backspace operation from the current cursor position. This action might do several things:

removing the current selection if available.

removing the previous char and move the cursor back.

do nothing, if we are at the start.

Example

main.py

from kivy.app import App
from kivy.uix.textinput import TextInput
import re


class DateInput(TextInput):

    def do_backspace(self, from_undo=False, mode='bkspc'):
        print(from_undo, mode)
        if len(self.text) >= 1:
            if self.text[-1] == "/":
                self.text = self.text[:-1]
                return True    # we have consumed the backspace and don’t want it to propagate any further
        return super(DateInput, self).do_backspace(from_undo, mode)

    #filter for date formatting which works well aside from backspacing
    pat = re.compile('[^0-9]')

    def insert_text(self, substring, from_undo=False):
        pat = self.pat
        if len(substring) > 1:
            substring = re.sub(pat, '', (self.text + substring))
            self.text = ''
            slen = len(substring)
            if slen == 2:
                s = substring[:2] + '/'
            elif slen == 3:
                s = substring[:2] + '/' + substring[2:]
            elif slen == 4:
                s = substring[:2] + '/' + substring[2:] + '/'
            else:
                s = substring[:2] + '/' + substring[2:4] + '/' + substring[4:8]
        elif len(self.text) > 9:
            s = ''
        elif len(self.text) == 1:
            s = re.sub(pat, '', substring)
            if s != '':
                s = s + '/'
        elif len(self.text) == 4:
            s = re.sub(pat, '', substring)
            if s != '':
                s = s + '/'
        else:
            s = re.sub(pat, '', substring)
        return super(DateInput, self).insert_text(s, from_undo=from_undo)


class TestApp(App):

    def build(self):
        return DateInput()


if __name__ == "__main__":
    TestApp().run()

Output

score 0 · Answer 4 · answered Apr 06 '21 at 04:26

Here is my solution:

class DateInput(TextInput):
    pat = re.compile("[^0-9]")

    def insert_text(self, substring, from_undo=False):
        pat = self.pat
        s = re.sub(pat, "", substring)
        if len(self.text) >= 10:
            return super(DateInput, self).insert_text("", from_undo=from_undo)
        elif ((len(self.text) > 2) and (self.cursor_index() == 2)) or (
            (len(self.text) > 5) and (self.cursor_index() == 5)
        ):
            return super(DateInput, self).insert_text("/", from_undo=from_undo)
        elif (len(self.text) == 2) or (len(self.text) == 5):
            return super(DateInput, self).insert_text("/" + s, from_undo=from_undo)
        else:
            return super(DateInput, self).insert_text(s, from_undo=from_undo)

Kivy Python: Detect backspace in Text Input

4 Answers4

Solution

Example

main.py

Output