正则表达式不匹配序列中的两个相同字符

Question

I know this is very simple for you guys, but i'm scratching my head to find the regular expression to match below strings. Have tried different reg ex, but failed every time. Below are the string that needs to be matched/unmatched.

S__ATHEESH – not match
S_A_T_HEESH – match 
S’_ATHEESH – match 
S’’ATHEESH – not match
S-A_THEESH - match

and here is the requirement

a.       A name can have the special characters – space, apostrophe, underscore and hyphen. 
b.      There can be more than 1 special character in a name, but same special character cannot repeat more than once continuously. 
c.       There should be minimum 2 characters entered in the name field 

I have below reg ex, that needs to be modified

^([a-zA-Z]+[ _'-])*[a-zA-Z]+$

Thanks for your help.

Answer 1

You can use a lookahead based solution like

/^(?!.*([-_ ’])\1)(?=(?:[^a-z]*[a-z]){2})[-_ ’a-z]+$/i

• (?!.*([-_ '])\\1) checks, that there is no doubled special character
• (?=(?:[^az]*[az]){2}) checks the presence of at least two letters
• ^[-_ 'az]+$ matches, if only the allowed characters are present

I used case-insenstive modifier to avoid writing a-zA-Z all the time. I'm not sure ' is the correct hyphen character, but it's the one you used.

Here's a demo: https://regex101.com/r/fD1gV0/1

Answer 2

The following regular expression will meet your requirements:

^(?=.{2})(?:[a-zA-Z]+|([ _’'-])(?!\1))+$

• ^ assert position at the beginning of the string
• (?=.{2}) at least 2 characters
• and (:
  • [a-zA-Z]+ at least one alphabetic character
• or:
  • ([ _''-]) a space, underscore, magic apostrophe, apostrophe or dash, captured in capture group 1
  • (?!\\1) negative lookahead that what was just captured isn't also the next character
• ) + as many times as possible
• $ assert position at the end of the string

Demo

Answer 3

Correct me if I'm wrong, but your requirement is simply for strings which do not contains 2 spaces (or apostrophes, underscores or hyphens) one after the other.

In RegEx land, AFAICT, it's not straightforward to assert a "not", but the requirement seems simple enough to be treated in reverse:

(  |__|''|’’)

will match what you mark as "not match", so you just have to negate the boolean match result.

Good luck.

Edit: just to make sure I'm not misunderstood. I'm not saying you cannot do exactly what the op is asking, I'm just saying that negating the requirement is a lot simpler and easier to maintain by someone else.