python - 如何修复函数tkinter的int float

Question

所以我试图创建一个函数，检查用户是否输入了str内容，它将打印出一条错误消息，说print("You can only use a number ranging from 1 - 4, setting minimum_offer to 1.")但是当我运行我的代码时，它说这是错误代码:

During handling of the above exception, another exception occurred:
Traceback (most recent call last):
  File "C:\Program Files\Python38\lib\tkinter\__init__.py", line 1883, in __call__
    return self.func(*args)
  File "d:/Python Code stuff I did/print.py", line 218, in <lambda>
    button = Button(root,text="Change Config", width=20, height=3, bg="#0f0f0f",fg="#ffffff", command=lambda:[do_it(), do_it1(), do_it2(), do_it3(),do_the_it(),do_the_it1()])
  File "d:/Python Code stuff I did/print.py", line 42, in do_it
    userInput = number.get()
  File "C:\Program Files\Python38\lib\tkinter\__init__.py", line 537, in get
    return int(self._tk.getdouble(value))
_tkinter.TclError: expected floating-point number but got "wa"

这是我的代码，给出错误:

number_1 = StringVar()
def do_the_it():
    userinput4 = number_1.get()
    try:
        int(userinput4)
        updater.read('settings.ini')
        updater['Trading Settings']['minimum_offer'].value = number_1.get()
        updater.update_file()
    except Exception as error:
        print("You can only use a number ranging from 1 - 4, setting minimum_offer to 1.")
        updater.read('settings.ini')
        updater['Trading Settings']['minimum_offer'].value = 1
        updater.update_file()

    updater.read('settings.ini')
    updater['Trading Settings']['maximum_offer'].value = number_2.get()
    updater.update_file()
    a = updater['Trading Settings']['minimum_offer'].value
    if int(number_1.get()) > 4:
        print("Number can only exceed to 4")
        updater.read('settings.ini')
        updater['Trading Settings']['minimum_offer'].value = 1
        updater.update_file()

Answer 1

为了进行错误处理，您必须检查是否确实可以转换为int。您可以使用try块。

accept = True
try:
    tmp = int(number_1.get())
except:
    accept = False

然后，如果accept为true，则只能执行第二部分(if块):

if accept:
    if int(number_1.get()) > 4:
        ### Rest of your code
else:
    print("Enter a number you joker")