Remove punctuation from beginning and end of words in a string

Question

I have a string and I want to remove any - ,( ,) ,& ,$ ,# ,! ,[ ,] ,{ ,} ," ,' from the beginning or end of the word. If it is between the words just ignore those.

Here is an example:

var $string = "Hello I am a string, and I am called Jack-Tack!. My -mom is Darlean.";

I want the above text to be like this after a regex:

console.log("Hello I am a string and I am called Jack-Tack My mom is Darlean");

your example removes the puntuation insides the string also, edit your title to "Remove punctuation from string with js" or edit the example. And jQuery is for DOM manipulation, not for data manipulation — Einacio
– Einacio, Commented Aug 18, 2011 at 16:00
It isn't really clear what "Hello!^ there" and "Hello^! there" should be converted into. My solution will convert the former into "Hello^ there" and leave the latter untouched where as Joseph Silber's solution will do the opposite (convert the latter into "Hello^ there" while leaving the former untouched). It all depends on what "a word" is for you. — Wladimir Palant
– Wladimir Palant, Commented Aug 18, 2011 at 16:51

Wladimir Palant · Accepted Answer · 2011-08-18 16:45:37Z

You can use the fact that \b in regular expressions always matches at a word boundary whereas \B matches only where there isn't a word boundary. What you want is removing this set of characters only when there is a word boundary on one side of it but not on the other. This regular expression should do:

var str = "Hello I'm a string, and I am called Jack-Tack!. My -mom is Darlean."; str = str.replace(/\b[-.,()&$#!\[\]{}"']+\B|\B[-.,()&$#!\[\]{}"']+\b/g, ""); console.log(str); // Hello I'm a string and I am called Jack-Tack My mom is Darlean

Joseph Silber · Accepted Answer · 2011-08-18 16:17:34Z

Here's one I just created. It could probably be simplified, but it works.

"Hello I am a string, and I am called Jack-Tack!. My -mom is Darlean." .replace(/(?:[\(\)\-&$#!\[\]{}\"\',\.]+(?:\s|$)|(?:^|\s)[\(\)\-&$#!\[\]{}\"\',\.]+)/g, ' ') .trim();

Jacob Eggers · Accepted Answer · 2011-08-18 16:56:14Z

Replace this regex with an empty string.

/(?<=\s)[^A-Z]+?(?=\w)|(?<=\w)[^a-z]*?(?=\s|$)/gi //Explanation: / (?<=\s)[^A-Z]+?(?=\w) //Find any non-letters preceded by a space and followed by a word | //OR (?<=\w)[^a-z]*?(?=\s|$) //Any non-letters preceded by a work and followed by a space. /gxi

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