Question

如果输入的是字符串，我想请用户输入一个值，然后打印其长度。如果用户输入的是整数或小数，我想分别打印"Sorry, integers don't have length"或"Sorry, floats don't have length"。

在尝试将输入转换为float或integer时，我正在使用异常。

这是我的代码：

c=input("enter a string: ")

def length(c):
    return len(c)

try:
    float(c)
    if float(c)==int(c):
        print("Sorry integers don't have length")
    else:
        print("Sorry floats don't have length")
except:
    print(length(c))

输出结果如下：

> enter a string: sfkelkrelte
11
> enter a string: 21
Sorry integers don't have length
> enter a string: 21.1
4

当我输入整数时，错误消息会正确显示，因为可以进行float（）转换。但是，在浮点数的情况下，解释器将转到except块，尽管它应该已经执行了try块。

为什么会这样？

Answer 1

还为什么不将EAFP原则（What is the EAFP principle in Python?）应用于第二种条件？

select
    count(*) as cnt
from yourTable
where [date] > (select max([date]) from yourTable where [status] = 1)

我已经省略了检查s = input("Input: ") try: float(s) try: int(s) print("Sorry, you have entered an integer.") except: print("Sorry, you have entered a float.") except: print(len(s))的例外情况，因为您也没有在代码中检查它。但是，您应该看看How to properly ignore exceptions

Answer 2

引发异常的部分是int(c)：

c = "21.1"
print(int(c))
# ValueError: invalid literal for int() with base 10: '21.1'

一个较小的更改将为您解决此问题：

c="21.1"

try:
    float(c)
    if "." not in c: #<---- check if "." is in the string c AFTER conversion to float
        print("Sorry integers don't have length")
    else:
        print("Sorry floats don't have length")
except:
    print(len(c))
# Sorry floats don't have length

但是，盲目地捕获所有异常通常是一种不好的做法。这会在程序中触发except块的错误中产生意外的副作用。

仅捕获您期望的那个会更合适。

c="21.aakdjs1"

try:
    float(c)
    if "." not in c:
        print("Sorry integers don't have length")
    else:
        print("Sorry floats don't have length")
except ValueError:
    print(len(c))
# 10

对于以后的调试，您随时可以print the exception。

Answer 3

将结果输入为字符串格式，因此将字符串转换为float，然后检查它是否为整数。将您的代码更改为：

old: if float(c) ==int(c):

new: if c.isdigit():

已更新：

enter a string: 21.0
Sorry floats don't have length

尝试将输入字符串转换为float

3 个答案: