Python密文,if else判断和while,for循环

1,编写用户名及密码然后加密

import getpass
username = input("username")
password = getpass.getpass("password")
print(username,password)

变秘文需要有模块,调用标准库(import中有的,不需要安装的库,叫做标准库)

import getpass,可以将明文变成密文

'''
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'''
import getpass
username = input("username")
password = getpass.getpass("password")
print(username,password)

2. 想要判断用户名和密码对不对,所以我们需要判断

username_ = "han"
password_ =  "123"
username = input("username:")
password = input("password:")
print(username,password)
 
if username_ == username and password_ == password:
    print("welcome  login...")
else:
    print("密码或者用户名错误.")

python中没有分隔符,所以有相应的强制缩进,省的结束语。如果总显示错误,写的代码没问题,一定要检查缩进问题。

3.判断年龄的大小

'''
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'''
age_ = 30
age = int(input("Enter your age:"))
if age == age_:
    print("you are right!")
elif age_ < age:
    print("It is high")
else :
    print("It is low!")

在python中 if-eiif-lese的循环结构较为简单,在循环体中记得要加“:”

4.while 循环

判断年龄想要加功能,让其可以猜想三次

while True:
    if count == 3:
        break
    age_ =30
    age = int(input("Enter your age:"))
    if age == age_:
        print("you are right!")
        break
    elif age_ < age:
        print("It is high")
    else:
        print("It is low!")
 
    count +=

优化一下:

'''
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寻找有志同道合的小伙伴,互帮互助,群里还有不错的视频学习教程和PDF电子书!
'''
 count = 0<br>while count < 3:
    age_ =30
    age = int(input("Enter your age:"))
    if age == age_:
        print("you are right!")
        break
    elif age_ < age:
        print("It is high")
    else:
        print("It is low!")
 
    count += 1
if count == 3#else:
print("你打印了太多次,请重新登陆!")

5.for循环

for i in range(10):
    print("loop",i)

想要跳一个打印一个

for i in range(0102):
    print("loop",i)

将while循环改为for循环

'''
遇到问题没人解答?小编创建了一个Python学习交流QQ群:579817333 
寻找有志同道合的小伙伴,互帮互助,群里还有不错的视频学习教程和PDF电子书!
'''
for i in range(3):
    age_ =30
    age = int(input("Enter your age:"))
    if age == age_:
        print("you are right!")
        break
    elif age_ < age:
        print("It is high")
    else:
        print("It is low!")

range()的用法见python的函数随笔

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转载自blog.csdn.net/sinat_38682860/article/details/105390297