JavaScript Regex exclude leading and trailing spaces, but maintain inner-spaces

Question

I have a string like this:

' hello world '

I would like to match groups ' ', 'hello world', and ' '

I can't figure out how to allow spaces in the middle of the regex, where they aren't anchored to end

My attempt (http://regex101.com) /(^\s*)(.+)(\s*$)/g

1. Problem is underspecified. What should happen with ' '? 2. Why do you need this in a single regex? — melpomene
– melpomene, Commented Oct 23, 2017 at 20:21
It is very strange what you need it for. Surely ^(\s*)(.*?)(\s*)$ will work, but it might slow your code down significantly. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Oct 23, 2017 at 20:22
I don't know your exact use case, but what about simply using the trim() function? developer.mozilla.org/de/docs/Web/JavaScript/Reference/… — obecker
– obecker, Commented Oct 23, 2017 at 20:22
The use-case is irrelevant for this question, I need to match anything between the bounding spaces, but not the bounding spaces, if they exist — neaumusic
– neaumusic, Commented Oct 23, 2017 at 20:23

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

You can use this regex with 3 capturing groups:

/^(\s*)(\S+(?:\s+\S+)*)(\s*)$/

RegEx Demo

RegEx Breakup:

^: Start
(\s*): Match & capture starting white space (zero or more)
(\S+(?:\s+\S+)*): Match & capture middle string that may contain white spaces
(\s*): Match & capture ending white space (zero or more)
$: End

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Oct 23, 2017 at 20:24

anubhava

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4 Comments

neaumusic Over a year ago

the middle part doesn't seem as concise as it should be, I'm trying to understand it

anubhava Over a year ago

It is not short because I made an attempt to make it efficient, without lot of backtracking.

neaumusic Over a year ago

ah, now I see why it's \s+\S+ because if there is a space, another non-space is required after it

anubhava Over a year ago

Yes that's right as middle part must start and end with a non-space

Killzone Kid · Accepted Answer · 2017-10-23 20:29:29Z

1

Assuming your string is always spaces_words_spaces:

"   hello world   ".match(/^(\s+)(.*?)(\s+)$/)

Prints:

["   hello world   ", "   ", "hello world", "   ", index: 0, input: "   hello world   "]

answered Oct 23, 2017 at 20:29

Killzone Kid

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4 Comments

anubhava Over a year ago

.*? or .+? may also work but will be very inefficient as the input string length grows.

Killzone Kid Over a year ago

.*? is non-greedy quantifier

neaumusic Over a year ago

I see, never used lazy vs greedy before, thought that was the same as .? but now I get it

neaumusic Over a year ago

I gave @anubhava the answer for performance and lack of back-tracking, but this is still good and I learned something, thank you

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