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4 简单的 if 语句

4 简单的 if 语句

作者: 陪伴_520 | 来源:发表于2018-12-28 14:43 被阅读0次
    #一个简单的if语句
    animals = ['dog', 'cat', 'pig', 'horse']
    for pet in animals:
        if pet == 'dog':
            print("My favourite animals is " + pet.upper())
        else:
            print("Not " + pet.title())
            
    print("---------------------------> 多个条件的情况")
    #也可以检查多个条件 ---- 用and连接
    i = 5
    love = 2 
    you = 0
    if i == 5 and love == 2 and you >= 0:
        print("I love you!")
        
    #也可以用 or 做 或 判断
    if i == 5 or love == 6 or you > 0:
        print("ditto.")
        
    print("---------------------------> 检查特定值是否在列表中")
    char = ['I', 'love', 'you', "hui"]
    #在表里的情况
    if 'hui' in char:
        print(char)
    #不在表里的情况
    if 'zong' not in char:
        print("Looking for...")
    #布尔语句本质上就是If语句,其只有两种状态 True 或 Flase
    
    number = 19
    number > 18  #python可以直接给出真假的判断
    
    #if else 和if elif else
    Pass = 60
    Good = 80
    Excellence = 90
    message = input("Please input your score:")
    score = int(message)
    if score >= Pass and score < 80:
        print("Luckily,you have passed!")
    elif score >= Good and score <= 90:
        print("Congratulation,You got a Good in the test!")
    elif score >= Excellence:
        print("NB! You got a Excellence!")
    else:
        print("Fight again...")
    

    皮一下😄

    #上海理工大学2019考研---控制工程之来不来复试自测。
    message = input("初试考了多少分啊?")
    score = int (message)          #注意要把输入转换成int类型
    if score >= 300 and score < 350:
        print("当然来复试!")
    elif score < 300 and score >= 270:
        print("那也得来复试!")
    elif score >= 350:
        print("滚...你不需要测试🙂")
    elif score < 270:              #最后一条或者直接是else,elif必须加上条件
        print("你已经很努力了,考研不是唯一出路......")
    
    #Pizza店 ---> 确认顾客是否要点普通Pizza(确认列表是否为空)
    requested_toppings = ['mushroom', 'extra cheese', 'lemon']
    ingredients = list(requested_toppings)
    
    print("There are available ingredients: " + str(ingredients))
    requested_topping = input("Please input the ingredients you want or input 0:")
    
    if requested_topping == '0':
        print("Are you sure you want a plain Pizza?")
        message = input("Y or N ? ")
        if message == "N":
            print("Please choose again!")
        if message == "Y":
            print("Thanks for your patience!")
    else:
        print("Adding " + requested_topping +".")
        print("\nFinshed your Pizza!")    
    
    #动手试一试 ---> 输出序数
    number = [1, 2, 3, 4, 5, 6, 7, 8, 9]
    for ordinal in number:
        if ordinal == 1:
            print(str(ordinal) + "st")
        elif ordinal == 2:
            print(str(ordinal) + "nd")
        elif ordinal == 3:
            print(str(ordinal) + "rd")
        else:
            print(str(ordinal) + "th")
    

    Date 2018年12月28日

    ac3b2ebb76ec8b60d79a995a4221b713.jpeg

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