如何在字符串中转换数字以完成python 2.7中的功能？

Question

我正在尝试使用函数编写一些代码，if / elif和循环。我将它基于学习Python的艰难之路，练习35.（Python 2,7）

我目前卡住的地方是def temp-function。当我输入数字时，我无法让程序接受用户输入。

我收到以下错误：

Traceback (most recent call last):
  File "ex35_1.py", line 53, in <module>
    temp ()
  File "ex35_1.py", line 11, in temp
    if number in next > 5:
TypeError: 'in <string>' requires string as left operand, not int

from sys import exit

def temp():
    print "Good morning."
    print "Let's get ready to kindergarden!"
    print "How cold is it outside?"

    #I think this is where the first problem is. 
        #The number-command is somehow wrong. 
    next = raw_input("> ")
    number = int(next)
    if number in next > 5:
        wool()
    elif number in next =< 6:
        not_wool()
    else:
        print "Fine, we just go!"

def wool():
    print "OK, so it is pretty cold outside!"
    print "Put on the wool."
    print "But is it raining?"
    rain = True

    while True:
        next = raw_input("> ")

        if next == "Yes":
            print "Put on the rain coat!"
            rain()
        elif next == "No" and rain:
            print "It is raining, but I dont wanna stress with the rain coat!"
            rain = False
        elif next == "No":
            print "You dont need a raincoat."
            march("With wool and no raincoat.")
        else:
            print "You should make a choice."
            exit(0)


def march(wit):
    print wit, "You are out the door!"
    exit (0)

def rain():
    print "Got the wellis?"
    march("With wool and rain stuff!")

def not_wool():
    print "There is no need for all that clothing."
    march("Remember the lunch pack!")

temp ()

有关上述错误的任何提示，以及可能的其他错误将不胜感激。

Answer 1

由于number已经是一个整数，你可以直接比较它。

number > 5

Answer 2

您转换为int并将其分配给变量编号。然后你尝试在原始字符串中找到int。

为什么需要在字符串中搜索int？你不能只评估int吗？

您可以通过在将输入分配给下一个时将输入转换为int来剪切一行。

另外= <应该是<=

def temp():
    print "Good morning."
    print "Let's get ready to kindergarden!"
    print "How cold is it outside?"

    next = int(raw_input("> "))

    if next  > 5:
        wool()
    elif next <= 6:
        not_wool()
    else:
        print "Fine, we just go!"