如何在Python 3.2中选择输入

当我运行该程序时，默认为喜剧，然后是Tartuffe。

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如何让它识别字符串中的不同类型？

您需要存储输入，然后将其与所需内容进行比较，例如：

input("Would you like to read: comedy, political, philisophical, or tragedy?")

a = "comedy"
b = "political"
c = "philisophical"
d = "tragedy"

if a:
    input("Would you like the author's nationality to be: English or French?")
    e = "French"
    d = "English"
    if e:
        print("Tartuffe")
    elif d:
        print("Taming of the Shrew")

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在我看来，更好的方法是这样做：

a = "comedy"
b = "political"
c = "philisophical"
d = "tragedy"

user_input = input("Would you like to read: comedy, political, philisophical, or tragedy?")

if user_input == a:
    user_input = input("Would you like the author's nationality to be: English or French?")

    if user_input == e:
        #do more stuff

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因为它更容易管理和读取不同的输入。

您只是在测试e的值是否为真（字符串不为空，因此为真）

您也没有存储输入

def comedy():
    print("comedy")

def political():
    print("political")

def philisophical():
    print("philisophical")

def tragedy():
    print("tragedy")

types = {"comedy":comedy,
         "political":political,
         "philisophical":philisophical,
         "tragedy":tragedy
        }

user_input = input()

types[user_input]()

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高度可扩展的代码

selection = input("Would you like the author's nationality to be: English or French? ")

if selection == e:
    print("Tartuffe")
elif selection == d:
    print("Taming of the Shrew")

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说得好。“你难道不喜欢一流的东西吗？”Makoto挑衅地说！这和函数指针（来自C++）使输入处理变得更加容易。@gnibbler谢谢，捕捉得好！我使用Python2.7进行测试，但忘了更改：）

choices = {'e': ('French', 'Tartuffe'), 'd': ('English', 'Taming of the Shrew')}

cstr = ', '.join('%r = %s' % (k, choices[k][0]) for k in sorted(choices))
prompt = 'What would you like the author\'s nationality to be (%s): ' % cstr

i = input(prompt).lower()

print('%s: %s' % choices.get(i, ('Unknown', 'Untitled')))