regex - 如何针对以下情况编写正则表达式(How do I write a regex for the following condition)

Question

• Should be at least 4 characters

  (至少应包含4个字符)

• Should start with a character [a-zA-Z]

  (应以字符[a-zA-Z]开头)

• Should not end with _

  (不应以_结尾)

• Should only contain at most 1 _ in the entire word

  (整个单词中最多只能包含1个_)

• The word can contain [a-zA-Z0-9]

  (该词可以包含[a-zA-Z0-9])

I tried the following regex:

(我尝试了以下正则表达式：)

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

But then I am wondering wether this could be made even smaller, and also I am not sure how do I set 'atleast 4 characters' constraint.

(但是然后我想知道是否可以使它更小，而且我不确定如何设置“至少4个字符”约束。)

  ask by Ranjandas translate from so

Answer 1

You can shorten your pattern by omitting the last negated character class [^_] as that will match any char other than _ and add a positive lookahead (?=.{4}) to assert 4 character from the start of the string:

(您可以通过省略最后一个否定的字符类[^_]来缩短您的模式，因为它将匹配_以外的其他任何字符，并添加正向超前(?=.{4})以从字符串开头声明4个字符：)

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

• ^ Start of string

  (^字符串开头)

• (?=.{4}) Assert 4 chars

  ((?=.{4}) 4个字符)

• [a-zA-Z] Match a single char a-zA-Z

  ([a-zA-Z]匹配一个字符a-zA-Z)

• [a-zA-Z0-9]*_?[a-zA-Z0-9]+ Match an optional _ and any of the listed on the left and/or right

  ([a-zA-Z0-9]*_?[a-zA-Z0-9]+匹配可选的_和左侧和/或右侧列出的任何一个)

• $ End of string

  ($字符串结尾)

Regex demo

(正则表达式演示)