Question

我正在制作一个程序，使用python 3.4确定车牌是否处于正确的顺序（我正在开始编程，并做一些自我指定的家庭工作）。

牌照应该是三个字母的顺序，三个数字是正确的顺序。

这是我的代码：

#Get data from user
plate = input('Enter the lisence plate number: ')

#Determine if style is old or new
if len(plate) == 6 and plate[0] >= "A" and plate[0] <= "Z"\
   and plate[1] >= "A" and plate[1] <= "Z"\
   and plate[2] >= "A" and plate[2] <= "Z"\
   and plate[3] >= "0" and plate[1] <= "9"\
   and plate[4] >= "0" and plate[4] <= "9"\
   and plate[5] >= "0" and plate[5] <= "9":
    verd = 'works'
else: 
    verd = 'Not work'

#print results
    print(verd)

当我输入牌照ABC123时，它告诉我它不起作用。

我一直在尝试一切，但无法弄清楚为什么这不起作用。

任何帮助都将不胜感激。

Answer 1

顺便说一句，你方法中的错误是第三种情况 -

awk '{print i, $1, $2, $3, $4}' ifile.txt > ofile.txt

将其更改为 -

and plate[3] >= "0" and plate[1] <= "9"   <-------- Notice that you are using `plate[1]` instead of [3]

Answer 2

一个简单的正则表达式将完成这项工作。

re.match(r'[A-Z]{3}[0-9]{3}$', s)

由于re.match尝试从字符串的开头匹配，因此您不需要使用行锚^的开头。

示例：

>>> import re >>> def check(s): if re.match(r'[A-Z]{3}[0-9]{3}$', s): print('works') else: print('Not work') >>> check(input('Enter the lisence plate number: ')) Enter the lisence plate number: ABC123 works >>> check(input('Enter the lisence plate number: ')) Enter the lisence plate number: ABCDEF Not work >>>

Answer 3

在线：

and plate[3] >= "0" and plate[1] <= "9"\

将'1'改为'3'，如下所示：

and plate[3] >= "0" and plate[3] <= "9"\

Answer 4

你的错误是一个错字。

   and plate[3] >= "0" and plate[1] <= "9"\

与

   and plate[3] >= "0" and plate[3] <= "9"\

此外，print(verd)的缩进已关闭;它位于else区块中，因此只有在许可证无效的情况下才会打印任何内容。

这里是您的代码，只需进行少量更改即可使其正常运行：

#Get data from user
plate = input('Enter the lisence plate number: ')

#Determine if style is old or new
if len(plate) == 6 and plate[0] >= "A" and plate[0] <= "Z"\
   and plate[1] >= "A" and plate[1] <= "Z"\
   and plate[2] >= "A" and plate[2] <= "Z"\
   and plate[3] >= "0" and plate[3] <= "9"\  # Fixed bug here
   and plate[4] >= "0" and plate[4] <= "9"\
   and plate[5] >= "0" and plate[5] <= "9":
    verd = 'works'
else:
    verd = 'Not work'

#print results
print(verd)  # Fixed bug here

但是，您的代码还有其他一些问题，因此这里有一个改进的版本：

# Get data from user
plate = input('Enter the license plate number: ')

# Determine if style is old or new
if (len(plate) == 6 and
    'A' <= plate[0] <= 'Z' and
    'A' <= plate[1] <= 'Z' and
    'A' <= plate[2] <= 'Z' and
    '0' <= plate[3] <= '9' and
    '0' <= plate[4] <= '9' and
    '0' <= plate[5] <= '9'):
    verd = 'Valid'
else:
    verd = 'Invalid'

print(verd)

你也可以这样做：

# Get data from user
plate = input('Enter the license plate number: ')

# Determine if style is old or new
if (len(plate) == 6 and
    plate[:3].isupper() and plate[:3].isalpha() and
    plate[3:].isdigit():
    verd = 'Valid'
else:
    verd = 'Invalid'

print(verd)

但一般来说，我会说最好/最干净的方法是正则表达式方法：

import re

plate = input('Enter the license plate number: ')

if re.match(r'^[A-Z]{3}[0-9]{3}$', plate):
    print('Valid')
else:
    print('Invalid')

if语句不读取字符串中的字符

4 个答案: