由逗号分隔的正则表达式重复模式

Question

I have a regular expression of the following in Javascript:

/\w{1,}\s*\w{1,}/  

This checks if a string has atleast two words greater than 1 character each.

Ex-

asd fgh - Valid  
a b d dfd - Valid  
xscxs - Invalid

I now have a new requirement that I have tried to implement, but cannot get right.

New requirement: Be able to have a comma separated list of the same type of input as before. Cannot end with a comma. Each item must be valid per the rules above.
If there are no comma then also its valid.
Also all characters are alphabets and no numbers/special characters

Valid:  HOH vfdffd,dsfds dfgd,UIU fgfd  
Valid:  JOI JOIO  
Invalid:  QASW fgdfg,  
Invalid:  sdfds,1234 dfgdfg  
Invalid:  JKJ,ABCD  

Answer 1

You may specify the pattern for the first requirement as [a-zA-Z]+(?:\\s[A-Za-z]+)+ to match 1+ letters chars and then match 1+ sequences of a whitespace + one or more word chars, and then just repeat the pattern inside another group with a comma:

/^[a-z]+(?:\s[a-z]+)+(?:,[a-z]+(?:\s[a-z]+)+)*$/i

See the regex demo (all \\s replaced with spaces in the demo since the input is one multiline string).

If multiple spaces are allowed, replace \\s with \\s+ .

Details :

• ^ - start of string
• [az]+ - 1+ letter
• (?:\\s[az]+)+ - 1 or more sequences of (ie a space is obligatory)
  • \\s - a whitespace (add + to match 1 or more occrurrences)
  • [az]+ - 1+ letters
• (?:,[az]+(?:\\s[az]+)+)* - zero or more sequences (ie a comma is optional) of
  • , - a comma
  • (?:\\s[az]+)+ - see above
• $ - end of string.

Answer 2

.regex(/^(?:(?![0-9]{4}|[a-zA-Z]{4})[a-zA-Z0-9]{4})(?:(?:\b\,)(?![0-9]{4}|[a-zA-Z]{4})[a-zA-Z0-9]{4})*$/)

Explanation:

(?:\b\,) -> Match with a , at the beginning of the string only if its preceded by a word boundary
(?:(?![0-9]{4}|[a-zA-Z]{4})[a-zA-Z0-9]{4}) -> Match a string with letter and digits only if dont have 4 digits ow 4 letters