如何为以下条件编写正则表达式

Question

• 应至少为 4 个字符
• 应该以字符[a-zA-Z]开头
• 不应以_结尾
• 整个单词最多只能包含 1 个_
• 单词可以包含[a-zA-Z0-9]

我尝试了以下正则表达式：

^[a-zA-Z][a-zA-Z0-9]*_?[a-zA-Z0-9]*[^_]$

但是后来我想知道这是否可以做得更小，而且我不确定如何设置“至少 4 个字符”约束。

Answer 1

您可以通过省略最后一个否定字符类[^_]来缩短您的模式，因为它将匹配除_之外的任何字符，并添加一个正向前瞻(?=.{4})以断言从字符串开头的 4 个字符：

^(?=.{4})[a-zA-Z][a-zA-Z0-9]*_?[a-zA-Z0-9]+$

• ^字符串开始
• (?=.{4})断言 4 个字符
• [a-zA-Z]匹配单个字符 a-zA-Z
• [a-zA-Z0-9]*_?[a-zA-Z0-9]+匹配一个可选的_和左边和/或右边列出的任何一个
• $字符串结尾

正则表达式演示

Answer 2

这可以完成以下工作：

^(?i)(?=\w{4,})[a-z]+_?[^\W_]+$

解释：

^                   # beginning of line
  (?i)              # case insensitive
  (?=\w{4,})        # positive lookahead, make sure we have 4 or more word character
  [a-z]+            # 1 or more alphabet
  _?                # optional underscore
  [^\W_]+           # 1 or more alphanum
$

演示和说明

Answer 3

验证可以通过积极和消极的前瞻组合来完成：

import re

tests = [
    'abc', # too short
    '_bcde', # starts with wrong character
    'abcd_', # last character is '_'
    'a_b_cd' # too many '_',
    'abc&cd', # illegal character '&'
    'ab_cd' # OK
]


regex = re.compile(r"""
    ^               # matches start of the line
    (?=.{4})        # positive lookahead: matches any 4 characters (string must be at least 4 characters long)
    (?=[a-zA-Z])    # positive lookahead: next character must be [a-zA-Z]
    (?!.*_$)        # negative lookahead: last character cannot be `_`
    (?!.*_.*_)      # negative lookahead: cannot match more than one `_`
    [a-zA-Z_]+      # looking for one or more of these
    $               # looking for the end of the string
""", re.X)

for test in tests:
    m = regex.match(test)
    print(test, 'Match' if m else 'No match')

印刷：

abc No match
_bcde No match
abcd_ No match
a_b_cdabc&cd No match
ab_cd Match

请参阅正则表达式演示