如何在 while 循环中使用 raw_input() [英] How to use raw_input() with while-loop

问题描述

只是想写一个程序来接收用户输入并将其添加到列表数字"中:

print "Going to test my Knowledge here"打印输入一个 1 到 20 之间的数字:"i = raw_input('>>')数字 = []而 1 <= i <= 20 :打印好的，将 %d 添加到数字集:" % inumbers.append(i)打印好的，现在设置的数字是:"，数字

但是当我执行程序时它只运行到 raw_input()

在这里测试我的知识输入 1 到 20 之间的数字:>>>4

我在这里遗漏了一些基本规则吗??

解决方案

raw_input 返回字符串而不是整数:

所以，

<预><代码>>>>1 <= "4" <= 20错误的

使用int():

i = int(raw_input('>> '))

仅使用 if，如果您只从用户那里获取一个输入:

如果 1 <= i <= 20 :打印好的，将 %d 添加到数字集:" % inumbers.append(i)打印好的，现在设置的数字是:"，数字

将 while 用于多个输入:

i = int(raw_input('>> '))数字 = []而 1 <= i <= 20 :打印好的，将 %d 添加到数字集:" % inumbers.append(i)i = int(raw_input('>> ')) #再次请求输入打印好的，现在设置的数字是:"，数字

Just trying to write a program that will take the users input and add it to the list 'numbers':

print "Going to test my knowledge here"
print "Enter a number between 1 and 20:"

i = raw_input('>> ')
numbers = []

while 1 <= i <= 20 :
    print "Ok adding %d to numbers set: " % i 
    numbers.append(i)

    print "Okay the numbers set is now: " , numbers

However when I execute the program it only runs to raw_input()

Going to test my knowledge here
Enter a number between 1 and 20:
>>> 4

Is there some fundamental rule I'm missing here??

解决方案

raw_input returns a string not an integer:

So,

>>> 1 <= "4" <= 20
False

Use int():

i = int(raw_input('>> '))

Use just if, if you're only taking a single input from user:

if 1 <= i <= 20 :
    print "Ok adding %d to numbers set: " % i 
    numbers.append(i)

    print "Okay the numbers set is now: " , numbers

Use while for multiple inputs:

i = int(raw_input('>> '))
numbers = []

while 1 <= i <= 20 :
    print "Ok adding %d to numbers set: " % i 
    numbers.append(i)
    i = int(raw_input('>> '))                   #asks for input again
print "Okay the numbers set is now: " , numbers

这篇关于如何在 while 循环中使用 raw_input()的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！