Regular Expression: match only non-repeated occurrence of a character

I need to find and replace all occurrences of apostrophe character in a string, but only if this apostrophe is not followed by another apostrophe.

That is

abc'def

is a match but

abc''开发者_StackOverflowdef

is NOT a match.

I've already composed a working pattern - (^|[^'])'($|[^']) but I believe it may be shorter and simpler.

Thanks,

Valery

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depends on your environment - if your environment supports lookahead and lookbehind, you can do this: (?<!')'(?!')

Ref: http://www.regular-expressions.info/lookaround.html

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I think your pattern is short and precise. You could be using negative lookahead/lookbehind, but they would make it a lot more complex. Maintainability is important.

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You'll have to be careful for an uneven number of apostrophes:

abc'''def

where you probably do want to replace the 3^rd one and leave the 1^st and 2^nd in there.

You can do that like this (assuming you already matched string literals and only want to replace the uneven numbered trailing apostrophe):

Search for the pattern:

(('')*)'

and replace it with

$1

which is group 1: the even numbered apostrophes (or no apostrophes at all).

I'm not sure what actual problem you're solving, but in case you're parsing/reading a CSV file, or a string that has the likes of CSV input, I highly recommend using a decent CSV parser. Almost all languages have them in some form or another.

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see here nagative lookahed q(?!u)

• (?=pattern) is a positive look-ahead assertion
• (?!pattern) is a negative look-ahead assertion
• (?<=pattern) is a positive look-behind assertion
• (?<!pattern) is a negative look-behind assertion

http://www.regular-expressions.info/lookaround.html

working DEMO