Question

我正在尝试存储变量的先前值，其变量已通过对话框更改，然后如果取消对话框并返回None，则将该变量的值转换回先前设置的值。例如，考虑：

letter_height = 30 letter_width = 50

我希望用户通过此更改值，这确实有效：

def NewLetterDimensions():
    global letter_height
    letter_height = (numinput("New Letter Height", "Please set the new letter height: ", minval = 10, maxval = 170))
    if letter_height == None:
        listen()

    global letter_width
    letter_width = (numinput("New Letter Width", "Please set the new    letter width: ", minval = 10, maxval = 170))
    if letter_width == None:
        listen()

这被称为：

onkey(NewLetterDimensions, "k")

但是，如果对话框被取消，我希望“if”语句能够返回存储的先前值，以便将变量更改为先前的值而不是None。那么，我将如何实现这一目标呢？所以，我想要这个：

def NewLetterDimensions():
    global letter_height
    letter_height = (numinput("New Letter Height", "Please set the new letter height: ", minval = 10, maxval = 170))
    if letter_height == None:
        [] <- # Return previously set letter height (whether from previous dialog or not)
        listen()

global letter_width
letter_width = (numinput("New Letter Width", "Please set the new    letter width: ", minval = 10, maxval = 170))
if letter_width == None:
    [] <- # Return previously set letter width (whether from previous dialog or not)
    listen()

是的，这是龟图形。

Answer 1

怎么样：

user_input = (numinput("New Letter Height", "Please set the new letter height: ",
                             minval=10, maxval=170))
letter_height = letter_height if user_input is None else user_input

仅当letter_height不是user_input时才会更改为None，否则会保留旧值。

BTW，始终将None与is测试进行比较PEP8建议：

与None等单身人士的比较应始终使用is或is not，而不是等同运算符。

存储并返回同一变量

1 个答案: