6.5-1Python之列表嵌套字典的使用

基于字典的客户信息管理系统，使用列表嵌套字典完成。

代码逻辑参考【5.5Python之列表的嵌套使用】

在此基础上做优化处理：

# 初始化客户信息--列表嵌套字典
customers = [
    {"name": "Alice", "age": 25, "email": "alice@example.com"},
    {"name": "Bob", "age": 30, "email": "bob@example.com"},
    {"name": "Charlie", "age": 35, "email": "charlie@example.com"}
]

while True:
    print("""--------------欢迎进入客户管理系统--------------
           1. 添加客户
           2. 删除客户
           3. 修改客户
           4. 查询一个客户
           5. 查询所有客户
           6. 退出
-------------------------------------------""")
    choice = input("请输入你的选择：")
    # 1. 添加客户
    if choice == "1":
        name = input("请填写姓名：")
        age = int(input("请填写年龄："))
        email = input("请填写邮箱：")
        # 将原来的列表改为字典
        new_customer = {
                        "name": name,
                        "age": age,
                        "email": email
                        }
        customers.append(new_customer)
        print(f"你已添加新的客户信息：{new_customer}")
    # 2. 删除客户
    elif choice == "2":
        if len(customers) == 0:
            print(f"暂无客户信息！")
        else:
            delete_customer = input("请填写需要删除的客户姓名：")
            # 将原来的列表的索引改为字典的key值
            for customerD in customers:
                if customerD["name"] == delete_customer:
                    customers.remove(customerD)
                    print(f"你已删除客户的信息：{delete_customer}")
                    break
                elif (customers[len(customers) - 1])["name"] == customerD["name"]:
                    print(f"查无此客户的信息：{delete_customer}")
    # 3. 修改客户
    elif choice == "3":
        if len(customers) == 0:
            print(f"暂无客户信息！")
        else:
            update_customer = input("请填写需要修改的客户姓名：")
            for customerD in customers:
                # 将原来的列表的索引改为字典的key值
                if customerD["name"] == update_customer:
                    name = input("请填写新的姓名：")
                    age = int(input("请填写新的年龄："))
                    email = input("请填写新的邮箱：")
                    # 将原来的列表改为字典
                    customerD.update({"name": name, "age": age, "email": email})
                    print(f"你已将客户:{update_customer}的信息修改为：{customerD}")
                    break
                elif (customers[len(customers) - 1])["name"] == customerD["name"]:
                    print(f"查无此客户的信息：{update_customer}")
    # 4. 查询一个客户
    elif choice == "4":
        if len(customers) == 0:
            print(f"暂无客户信息！")
        else:
            search_customer = input("请填写需要查询的客户姓名：")
            for customerD in customers:
                # 将原来的列表的索引改为字典的key值
                if customerD["name"] == search_customer:
                    print(f"你查询的客户姓名: {customerD["name"]}, 年龄: {customerD["age"]}, 邮箱: {customerD["email"]}")
                    break
                elif (customers[len(customers) - 1])["name"] == customerD["name"]:
                    print(f"查无此客户的信息：{search_customer}")
    # 5. 查询所有客户
    elif choice == "5":
        if len(customers) == 0:
            print(f"暂无客户信息！")
        else:
            print(f"客户信息如下：")
            for customerD in customers:
                # 将原来的列表的索引改为字典的key值
                print(f"姓名：{customerD["name"]}，年龄：{customerD["age"]}，邮箱：{customerD["email"]}")
    # 6. 退出
    elif choice == "6":
        print("你已退出系统，感谢你的使用！")
        break
    # 其他情况
    else:
        print("输入的参数有误，请重新输入！")

代码逻辑上只是把原来的列表里的列表的逻辑修改为字典的逻辑。