正则表达式：如果第一个字符是 '%' 或字母数字然后

Question

in nodejs, I have to following pattern

(/^(>|<|>=|<=|!=|%)?[a-z0-9 ]+?(%)*$/i

to match only alphanumeric strings, with optional suffix and prefix with some special characters. And its working just fine.

Now I want to match the last '%' only if the first character is alphanumeric (case insensitive) or also a %. Then its optionally allowed, otherwise it should not match.

Example:

Should match:

>test
!=test
<test
>=test
<=test
%test
test
%test%
test%

Example which should not match:

<test% <-- its now matching, which is not correct
<test< <-- its now **not** matching, which is correct

Any Ideas?

Answer 1

You can add a negative lookahead after ^ like

/^(?![^a-z\d%].*%$)(?:[><]=?|!=|%)?[a-z\d ]+%*$/i
  ^^^^^^^^^^^^^^^^^  

See the regex demo . Details :

• ^ - start of string
• (?![^az\\d%].*%$) - fail the match if there is a char other than alphanumeric or % at the start and % at the end
• (?:[><]=?|!=|%)? - optionally match < , > , <= , >= , != or %
• [az\\d ]+ - one or more alphanumeric or space chars
• %* - zero or more % chars
• $ - end of string

Answer 2

You might use an alternation | to match either one of the options.

^(?:[a-z0-9%](?:[a-z0-9 ]*%)?|(?:[<>]=?|!=|%)?[a-z0-9 ]+)$

• ^ Start of string
• (?: Non capture group
  • [a-z0-9%] Match one of the listed in the character class
  • (?:[a-z0-9 ]*%)? Optionally match repeating 0+ times any of the character class followed by %
  • | Or
  • (?:[<>]=?|!=|%)? Optionally match one of the alternatives
  • [a-z0-9 ]+ Match 1+ times any of the character class
• ) Close non capture group
• $ End of string

Regex demo