当我只需要一个时，从我的字典中获得多个输出

Question

我只打印存储在字典中的相应缩写有一点问题

user = input("Enter a Abbreviation: ")

dictionary = {"ADSL": "Application Programming Interface",
              "IDE": "Integrated Development Enviroment",
              "SDK": "Software Development Kit",
              "UI": "User Interface",
              "UX": "User eXperience",
              "OPP": "Object Oriented Programming"
              }

for x in dictionary:
    if user in x:
        print(user + ":" + " " + dictionary[x])

    elif x != dictionary:
        print("Abbreviation not found") 

这是我的输出

Enter a Abbreviation: UI
Abbreviation not found
Abbreviation not found
Abbreviation not found
UI: User Interface
Abbreviation not found
Abbreviation not found 

我只需要输入缩写键值，不需要所有缩写没有找到输出，我希望这是有意义的。

Answer 1

字典的优点是它是“索引的”，所以你不需要找到关键。您遵循的是数组方法，这是一个错误的方法。

user = input("Enter a Abbreviation: ")

dictionary = {"ADSL": "Application Programming Interface",
              "IDE": "Integrated Development Enviroment",
              "SDK": "Software Development Kit",
              "UI": "User Interface",
              "UX": "User eXperience",
              "OPP": "Object Oriented Programming"
              }

if user in dictionary:
   print(user + ":" + " " + dictionary[user])
else:
   print("Abbreviation not found")

您可以找到更多关于字典这里、这里和google的信息。知道何时使用数组、字典、列表、队列非常重要.

Answer 2

也许只是这个：

user = input("Enter a Abbreviation: ")

dictionary = {"ADSL": "Application Programming Interface",
              "IDE": "Integrated Development Enviroment",
              "SDK": "Software Development Kit",
              "UI": "User Interface",
              "UX": "User eXperience",
              "OPP": "Object Oriented Programming"
              }

for x in dictionary:
    if user in x:
        print(user + ":" + " " + dictionary[x])
    else:
        continue

Answer 3

我试过这个密码，它有效..。看来你在检查所有的选项..。

user = input("Enter a Abbreviation: ")

dictionary = {"ADSL": "Application Programming Interface",
              "IDE": "Integrated Development Enviroment",
              "SDK": "Software Development Kit",
              "UI": "User Interface",
              "UX": "User eXperience",
              "OPP": "Object Oriented Programming"
              }

found = 0
for x in dictionary:
    if user in x:
        found = 1
        print(user + ":" + " " + dictionary[x])

if found == 0:
    print("Abbreviation not found")