Question

大家好，我希望你能解决一个我还没想到的小问题。这是对作业的描述，首先要了解我正在做的工作。

“在这部分作业中，你将编写一个输入两个不同的程序字符形式为“c1-c2”，其中c1 是第一个字符和c2 是第二个字符。您需要检查此输入的有效性。也就是说，输入应该以字母开头，后跟短划线（ - ），它应以另一个字母结尾不敏感（这意味着“a”和“A”是相同的字符）。请注意要从用户获得的字符必须彼此不同。在如果用户输入无效输入，您的程序应该连续进行提示所需的值，直到他输入有效值。“

以下是一些示例运行：

请在c1-c2表格中输入两个不同的字符：a-A
输入无效！
请输入两个不同的字符，格式为c1-c2：a-B （如果程序正确，程序应该停止。）

这是我到目前为止所尝试的内容：

Tensor

Answer 1

此处您的流程几乎总是在第一次检查时终止：

def process(Z):
    if Z[0] in ascii_letters:
        return True

您需要嵌套您的条件，并且仅在填写所有条件时“返回”，这是一个快速示例：

def process(Z):
    if Z[0] in ascii_letters:
        if Z[1] == '-':
           if Z[2] in ascii_letters:
              return True
    return False

请注意，在访问Z [2]之前，您需要检查Z的大小是否合适。您还需要添加有关同一字母的支票。因此，解决您的问题的方法是：

def process(Z):
    if len(Z) == 3
       and Z[0] in ascii_letters
       and Z[1] == '-'
       and Z[2] in ascii_letters
       and Z[0].upper() != Z[2].upper():
              return True
    return False

Answer 2

字符串具有可用于验证输入的方法。使用它们！您可以拆分＆＃39; - ＆＃39;并验证您有2个长度为1的字符，检查它们是否是字母并比较每个字符的大写版本。

Caused by: java.lang.BootstrapMethodError: java.lang.LinkageError: loading constraint violation when overriding method "org/hibernate/tool/schema/spi/DelayedDropRegistry.registerOnCloseAction(Lorg/hibernate/tool/schema/spi/DelayedDropAction;)V" during creation of class "org/hibernate/internal/SessionFactoryImpl$$Lambda$207/000000001B931600": loader "java/lang/InternalAnonymousClassLoader@1337bd9b" of class "org/hibernate/internal/SessionFactoryImpl$$Lambda$207/000000001B931600" and loader "com/ibm/ws/classloader/CompoundClassLoader@d5b1949f" of class "org/hibernate/tool/schema/spi/DelayedDropRegistry" have different types for the method signature
    at org.hibernate.internal.SessionFactoryImpl.<init>(SessionFactoryImpl.java:312)
    at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:460)
    at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:710)
    at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:726)
    at org.springframework.orm.hibernate5.LocalSessionFactoryBean.buildSessionFactory(LocalSessionFactoryBean.java:535)
    at org.springframework.orm.hibernate5.LocalSessionFactoryBean.afterPropertiesSet(LocalSessionFactoryBean.java:519)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1761)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1698)
    ... 156 more
Caused by: java.lang.LinkageError: loading constraint violation when overriding method "org/hibernate/tool/schema/spi/DelayedDropRegistry.registerOnCloseAction(Lorg/hibernate/tool/schema/spi/DelayedDropAction;)V" during creation of class "org/hibernate/internal/SessionFactoryImpl$$Lambda$207/000000001B931600": loader "java/lang/InternalAnonymousClassLoader@1337bd9b" of class "org/hibernate/internal/SessionFactoryImpl$$Lambda$207/000000001B931600" and loader "com/ibm/ws/classloader/CompoundClassLoader@d5b1949f" of class "org/hibernate/tool/schema/spi/DelayedDropRegistry" have different types for the method signature
    at sun.misc.Unsafe.defineAnonymousClass(Native Method)
    at java.lang.invoke.InnerClassLambdaMetafactory.spinInnerClass(InnerClassLambdaMetafactory.java:339)
    at java.lang.invoke.InnerClassLambdaMetafactory.buildCallSite(InnerClassLambdaMetafactory.java:206)

def process(Z): parts = Z.split("-") return (len(parts)==2 and len(parts[0])==1 and len(parts[1])==1 and parts[0].isalpha() and parts[0].upper() == parts[1].upper())适用于所有unicode字符，因此您无需担心用户是否以某种奇怪的语言输入文字。

Answer 3

处理此问题的正确方法是使用regular expressions。它们是一个正则表达式库，作为标准python库的一部分。

import re
u_input = 'your string'

def validate(input):
    # chars must be different
    if input[0] == input[-1:]:
        return False
    # begin and end with upper or lower a-z with dash in middle
    pattern = '[a-zA-Z]{1}-[a-zA-Z]{1}'
    result = re.match(pattern, input)
    # re.match returns None if no match
    if not result:
        return False:
    else:
        return True

Answer 4

我尝试使其尽可能与您的代码类似，并且理解起来非常简单。我想我考虑了所有必要的条件。如果没有，请告诉我。

from string import ascii_letters

def validity_check(input_chars):
    try:
        # if cant be splitted raises Exception which results in restart
        char_1, char_2 = input_chars.split("-")

        # if characters not in alphabet restarts
        chars_in_alphabet = char_1 in ascii_letters and char_2 in ascii_letters
        if not chars_in_alphabet:
            return False

        # if characters are the same letter restarts
        same_latter = char_1.lower() == char_2.lower()
        if same_latter:
            return False

        # I'm not sure if you want to check if the first letter
        # is small and the second one is capital
        # if so add this block
        # chars_letter_size = char_1 in ascii_letters.lower() and char_2 in ascii_letters.upper()
        # if not chars_letter_size:
        #     return False

        return True

    except:
        return False


while True:
    ask = input("Please enter two different characters in the form c1-c2: ")
    if validity_check(ask):
        break

Answer 5

内置功能 -isalpha（）将有助于您的作业。

while True:
    ask=input('Please enter two different characters in the form c1-c2: ')
    if len(ask)!=3:
        continue
    if ask[1] == '-' and ask[0].isalpha() and ask[2].isalpha() and ask[0]!=ask[2]:
        break

如果给定的字符串包含所有字母， isalpha（）函数将返回 True 。

如何使Python区分上下字符。

5 个答案: