Adding removing leading and trailing spaces in this JavaScript RegEx

Question

How do I add code to remove leading and trailing spaces to this regular expression?

I tried putting the \s in several places, but end up with odd results.

The myString is just the way it is, and is sent from a PHP script with trailing spaces.

Original Code

var myString = "::This is string 1 ::This is string 2! ";

myString = myString.replace(/\::(.+?)(?![^::])/g,'<li>$1</li>');

alert(myString);

Tried

var myString = "::This is string 1 ::This is string 2! ";

myString = myString.replace(/\::(.+?)(?![^::\s])/g,'<li>$1</li>');

alert(myString);

The end result I'm trying to achieve is

<li>This is string 1</li> // No trailing spaces before or after the `This` and `1`
<li>This is String 2</li>

Fiddle

Try ::(.+?)\s*(?=::|$)

Wiktor Stribiżew
– Wiktor Stribiżew

2016-02-04 08:24:15 +00:00
Commented Feb 4, 2016 at 8:24 — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Feb 4, 2016 at 8:24
trim

Tushar
– Tushar

2016-02-04 08:24:32 +00:00
Commented Feb 4, 2016 at 8:24 — Tushar
– Tushar, Commented Feb 4, 2016 at 8:24

Wiktor Stribiżew · Accepted Answer · 2016-02-04 08:30:39Z

4

The point is that you match spaces with .+?, and (?![^::\s]) only tells the regex engine to not match a character if it is followed by a whitespace or :.

Since you already are using lazy matching, you just need to use a greedy \s* subpattern to match whitespaces after the .+?.

Use

::(.+?)\s*(?=::|$)

See demo

Explanation:

:: - match 2 :s
(.+?) - match and capture into Group 1 one or more characters other than a newline
\s* - greedily match zero or more whitespaces
(?=::|$) - only if followed by :: or end of string.

And here is my attempt at unrolling the regex (looks a bit more efficient than the above):

::(\S*(?:(?=(\s+))\2(?!:|$)\S*)*)

See another demo

answered Feb 4, 2016 at 8:30

Wiktor Stribiżew

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Comments

Vegeta · Accepted Answer · 2016-02-04 08:56:41Z

1

Try this:

var myString = "::This is string 1 ::This is string 2! ";

myString = myString.replace(/\s*\::\s*(.*?)\s*(?=(::|$))/g,'<li>$1</li>');

JSFiddle

answered Feb 4, 2016 at 8:56

Vegeta

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