Python 3.4用户输入

Question

我想写一个小脚本，告诉我低音水平是否可以从用户输入。

我只是在学习用户输入，这就是我目前所拥有的：

def crisp():
    bass = input("Enter bass level on a scale of 1 to 5>>")
    print ("Bass level is at") + bass
    if bass >=4:
       print ("Bass is crisp")    
    elif bass < 4:
       print ("Bass is not so crisp")

Answer 1

转换为整数：

bass = int(input("Enter bass level on a scale of 1 to 5>>"))

Answer 2

当您通过内置函数接收input()时，它将输入作为字符串。

>>> x = input('Input: ')
Input: 1
>>> x
"1"

相反，将int()转换为input()：

>>> x = int(input('Input: '))
Input: 1
>>> x
1

否则，在您的代码中，您将检查if "4" == 4:，这永远不是真的。

因此，这里是您编辑的代码：

def crisp():
    bass = int(input("Enter bass level on a scale of 1 to 5>>"))
    print ("Bass level is at") + bass
    if bass >=4:
       print ("Bass is crisp")    
    elif bass < 4:
       print ("Bass is not so crisp")

Answer 3

我真的看不出有什么问题，但只是一个简单的程序来实现这一点，只有这样：

a=1
while a==1:
    try:
        bass = input('Enter Bass Level: ')
        print('Bass level is at ' + str(bass))
        if bass >=4:
            print("Bass is crisp")
        elif bass < 4:
            print('Bass is not so crisp')
        a=0
    except ValueError:
        print('Invalid Entry')
        a=1

与函数没有太大区别：

def Bass():
    a=1
    while a==0:
        try:
            bass = input('Enter Bass Level: ')
            print('Bass level is at ' + str(bass))
            if int(bass) >=4:
                print("Bass is crisp")
            elif bass < 4:
                print('Bass is not so crisp')
            a=0
    except ValueError:
        print('Invalid Entry')
        a=1