Question

我正在尝试从图书作业中编写一个小程序，但我无法检测用户的输入是int/float（增加到总数）还是string （return error）。我尝试在 add_to_total 变量上使用 .isdigit（）但是当我键入float时，它会直接跳到else代码块。我试过在互联网上搜索，但无法找到明确的答案。这是我的代码：

total = 0
print("Welcome to the receipt program!")

while True:
    add_to_total = raw_input("Enter the value for the seat ['q' to quit]: ")
    if add_to_total == 'q':
        print("*****")
        print "Total: $%s" % total
        break
    if add_to_total.isdigit(): #Don't know how to detect if variable is int or float at the same time.
        add_to_total = float(add_to_total)
        total += add_to_total
    else:
        print "I'm sorry, but '%s' isn't valid. Please try again." % add_to_total

任何答案都将不胜感激。

Answer 1

使用例外来捕获无法概括的用户输入。从测试任何用户输入切换到仅保护数学运算来总结收据。

total = 0
print("Welcome to the receipt program!")

while True:
    add_to_total = raw_input("Enter the value for the seat ['q' to quit]: ")
    if add_to_total == 'q':
        break
    try:
        total += float(add_to_total)
    except ValueError:
        print "I'm sorry, but '%s' isn't valid. Please try again." % add_to_total
print("*****")
print "Total: $%s" % total

Answer 2

您始终可以使用try... except方法：

try:
    add_to_total = float(add_to_total)
except ValueError:
    print "I'm sorry, but '%s' isn't valid. Please try again." % add_to_total
else:
    total += add_to_total

请记住：it's easier to ask forgiveness than permission

Answer 3

非常接近旧条目：How can I check if my python object is a number?。答案是：

isinstance(x, (int, long, float, complex))

Python：如果变量是整数或浮点数，则为真

3 个答案: