Kivy Python:检测文本输入中的退格
Kivy Python:检测文本输入中的退格
提问于 2018-06-29 11:08:10
我正在尝试为日期创建一个简单的TextInput，它将输入限制为数字，并自动填充(mm/dd/yy)格式的正斜杠。我成功地创建了一个过滤器，它通过重新定义insert_text()来实现这一点，除了当用户使用退格键时，我还想自动删除斜杠。但我不知道如何检测用户何时在文本输入中退格，以便在必要时触发一个事件来擦除斜杠。
下面的代码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？或者有更好的方法来做这件事？
# .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)回答 3
Stack Overflow用户
回答已采纳
发布于 2018-06-29 11:45:57
由于1.9.0版TextInput是FocusBehavior固有的，因此如果您想要检测何时按backspace，则必须使用keyboard_on_key_down()方法或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()Stack Overflow用户
发布于 2018-07-01 06:38:29
对于任何想要自动格式化日期输入的人来说，这是一个完整的解决方案：
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)Stack Overflow用户
发布于 2021-04-06 12:26:39
以下是我的解决方案：
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)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/51093710
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