为什么我的代码一直打印数组中的第一个数据？

Question

i = 0
one_adult = [20.00,30.00]
one_child = [12.00,18.00]
one_senior = [16.00,24.00]
family_ticket = [60.00,90.00]
groups_of_six = [15.00,22.50]
lion_feeding = 2.50
penguin_feeding = 2.00
evening_barbecue = 5.00


Ticket_Type = ["one_adult",one_adult[i],"one_child",one_child[i],"one_senior",one_senior[i],"family_ticket",family_ticket[i],"groups_of_six",groups_of_six[i]]


Attractions_1day = ["lion_feeding",lion_feeding,"penguin_feeding",penguin_feeding]
Attractions_2day = ["lion_feeding",lion_feeding,"penguin_feeding",penguin_feeding,"evening_barbecue",evening_barbecue]


option = input("are you buying tickets for one day? (yes or no)")
if option == "yes" :
  print()
  print("These are ticket types available for one day tickets and their prices $")
  print()
  print(Ticket_Type)
  print()
  print("These are the attractions available for one day tickets and their prices $")
  print()
  print(Attractions_1day)
elif option == "no" :
  i = i + 1
  print()
  print("These are ticket types available for two day tickets and their prices $")
  print()
  print(Ticket_Type)
  print()
  print("These are the attractions available for two day tickets and their prices $")
  print()
  print(Attractions_2day)

为什么它一直打印 i = 0 时的数据而不是 i = 1 时的数据？ 它应该打印 one_adult，30 而不是 one_adult 20 等该程序应该显示可供人们购买门票的选项（1 天/2 天）

Answer 1

您正在尝试在elif<\/code>中增加i<\/code> ； 但是，这不会影响Ticket_Type<\/code>的元素。

解决此问题的最快方法是在elif<\/code>语句中重新分配Ticket_Type<\/code>列表：

i = 0
one_adult = [20.00,30.00]
one_child = [12.00,18.00]
one_senior = [16.00,24.00]
family_ticket = [60.00,90.00]
groups_of_six = [15.00,22.50]
lion_feeding = 2.50
penguin_feeding = 2.00
evening_barbecue = 5.00


Ticket_Type = ["one_adult",one_adult[i],"one_child",one_child[i],"one_senior",one_senior[i],"family_ticket",family_ticket[i],"groups_of_six",groups_of_six[i]]


Attractions_1day = ["lion_feeding",lion_feeding,"penguin_feeding",penguin_feeding]
Attractions_2day = ["lion_feeding",lion_feeding,"penguin_feeding",penguin_feeding,"evening_barbecue",evening_barbecue]


option = input("are you buying tickets for one day? (yes or no)")
if option == "yes" :
  print()
  print("These are ticket types available for one day tickets and their prices $")
  print()
  print(Ticket_Type)
  print()
  print("These are the attractions available for one day tickets and their prices $")
  print()
  print(Attractions_1day)
elif option == "no" :
  i = i + 1
  Ticket_Type = ["one_adult",one_adult[i],"one_child",one_child[i],"one_senior",one_senior[i],"family_ticket",family_ticket[i],"groups_of_six",groups_of_six[i]]
  print()
  print("These are ticket types available for two day tickets and their prices $")
  print()
  print(Ticket_Type)
  print()
  print("These are the attractions available for two day tickets and their prices $")
  print()
  print(Attractions_2day)

Answer 2

移动线：

Ticket_Type = ["one_adult",one_adult[i],"one_child",one_child[i],"one_senior",one_senior[i],"family_ticket",family_ticket[i],"groups_of_six",groups_of_six[i]]

Answer 3

正确检查您的代码，您已指定i=0<\/code> ，并且在 Ticket_type 数组中的每个位置都放置了one_adult[i]<\/code> ，它只是数组的第一个元素，即“20”。 您没有在每次迭代后增加 i 的值。 就像你必须放置循环来增加值，以便打印数组的其他元素。

要解决您的问题，请执行此操作

["one_adult",one_adult[0],"one_child",one_child[1],"one_senior",one_senior[2],"family_ticket",family_ticket[3],"groups_of_six",groups_of_six[4]]