正则表达式在开始和结束时排除字符

Question

I'm trying to select all [a-z',] symbols but exclude commas at start and end. Keep only the selection inside the red rect. How can I do this?

Answer 1

I think what you want is an optional capturing group repeating zero or more times:

[a-z](?:[',]*[a-z]+)*

See the Online Demo

• [az] - A character in the range az.
• (?: - Open non-capturing group.
  • [',]* - Zero or more characters from the specified character class.
  • [az]+ - At least one character in the range az.
  • )* - Close non-capturing group and match it zero or more times.

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^{Note: If for good reasons OP's intention was to match a quote at the very end, we can add an optional apostrophe: [az](?:[',]*[az]+'?)* . See an online demo}

Answer 2

You could first replace the commas in question and select the characters you want afterwards:

 let string = `,,."?,tats,t'ats,t,s',l,f,%$;`. string = string,replace(/^,|,$/g; "~ooo~"). console;log(string);

In other languages you could use lookarounds (a negative lookbehind and a positive lookahead, that is).

Answer 3

You could match with the following regular expression.

[a-z'](?:[a-z',]*[a-z'])?

Start your engine!

Javascript's regex engine performs the following operations.

[a-z']      : match one character in character class
(?:         : begin non-capture group
  [a-z',]*  : match 0+ characters in character class
  [a-z']    : match one character in character class
)           : end non-capture group
?           : optionally match non-capture group