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，以将输入限制为数字，并自动填充（mm / dd / yy）格式的正斜杠。我成功创建了一个通过重新定义insert_text（）来实现此目的的过滤器，但当用户退格时，我也想自动删除斜杠。但是我不知道如何检测文本输入中用户何时退格，以便在必要时触发事件以消除斜杠。

这是一个片段，解释了我想要做什么，但是TextInput没有“ on_key_up”属性。有没有办法添加一个？还是解决这个问题的更好方法？

# .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

由于1。TextInput版本是FocusBehavior固有的，因此，如果要检测何时按退格键，必须使用keyboard_on_key_down()方法或keyboard_on_key_up()方法：< / p>

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

这里是想要自动格式化日期输入的任何人的完整解决方案：

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

在TextInput中重写do_backspace()方法。如果文本为“ /”，则返回True，表示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'/'，我们使用super（…）调用原始事件并返回结果。这样可以使do_backspace事件传播继续正常进行。

请参阅示例。

Text Input » do_backspace() method

do_backspace(from_undo=False, mode='bkspc')

     

从当前光标位置开始退格操作。这个动作   可能会做几件事：

     

  
• 删除当前选择（如果可用）。
  
• 删除上一个字符并将光标移回。
  
• 如果我们刚开始，什么也不要做。
  

示例

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)