Kivy Python：检测文本输入中的退格键

Question

我正在尝试为日期创建一个简单的 TextInput，将输入限制为数字并自动填充 (mm/dd/yy) 格式的正斜杠。我成功地创建了一个通过重新定义 insert_text() 来执行此操作的过滤器，除非用户退格，我也想自动删除斜杠。但是我不知道如何检测用户何时在文本输入中退格，以便我可以在必要时触发一个事件来擦除斜杠。

这是一个解释我想要做什么的片段，但 TextInput 没有 "on_key_up" 属性。有没有办法加一个？或者更好的方法来解决这个问题？

# .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)

Answer 1

自版本 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() 方法：

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()

Answer 2

这里是所有想要自动格式化日期输入的人的完整解决方案：

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)

Answer 3

解决方案

覆盖 TextInput 中的 do_backspace() 方法。如果文本为“/”，则 return 为真，表明我们已经消耗了退格键并且不希望它进一步传播。

最后，如果文本不是“/”，我们将使用 super(…) 调用原始事件并 return 结果。这允许 do_backspace 事件传播继续正常发生。

详情请参考示例

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.

例子

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()

输出

Answer 4

这是我的解决方案：

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：检测文本输入中的退格键

Kivy Python: Detect backspace in Text Input

python

keystroke

detection

backspace

kivy

解决方案

例子

main.py

输出