Javascript regex - specific number of characters in unordered string

Question

I'm trying to test whether or not an unordered string has '3' in it 5 times.

For example:

var re = /3{5}/;
re.test("333334"); //returns true as expected
re.test("334333"); //returns false since there is no chain of 5 3s

What regex would make the second line return true? If regex is not the best way to test this, what is?

Thanks!

Oriol · Accepted Answer · 2014-09-07 23:47:27Z

6

Try

(str.match(/3/g) || []).length >= 5

Or

str.split(3).length > 5

Where str is the string you want to test.

edited Sep 7, 2014 at 23:47

answered Sep 7, 2014 at 23:30

Oriol

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4 Comments

Sasha Chedygov Over a year ago

It should be str.split('3') (with the quotes). Fixed for you.

Oriol Over a year ago

@SashaChedygov Quotes are not necessary. From ES5 spec Section 15.5.4.14, at step 8, "If separator is a RegExp object (its [[Class]] is "RegExp"), let R = separator; otherwise let R = ToString(separator)."

Sasha Chedygov Over a year ago

@Oriol: Thanks for the clarification, I didn't know that. I'd still prefer the quotes, but that's just me. :)

Oriol Over a year ago

@SashaChedygov Also note that the split way needs > instead of >=, because it counts an additional item: ''.split(3).length === 1, '3'.split(3).length === 2, and so on.

Casimir et Hippolyte · Accepted Answer · 2014-09-07 23:31:06Z

3

You can write this:

var re = /(?:3[^3]*){5}/;

answered Sep 7, 2014 at 23:31

Casimir et Hippolyte

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2 Comments

Ativ Over a year ago

Hi, I'm new to regexp and am having trouble understanding capture parentheses. According the the MDN, this would match '3[^3]*' but not remember the match. What exactly does this mean?

Casimir et Hippolyte Over a year ago

@Ativ: yes (?:...) are non-capturing parenthesis. The goal is only to group, to be able to put the {5} quantifier in factor. A capturing group will store the content to be reuse later (with a backreference). But it is not needed here. Using non-capturing group use obviously less memory, and is a little faster.

bbonev · Accepted Answer · 2014-09-07 23:37:48Z

2

I would go for

s.replace(/[^3]/,'').length >= 5

Assuming that the string to be tested is named s

answered Sep 7, 2014 at 23:37

bbonev

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Comments

Lucas T. · Accepted Answer · 2014-09-08 16:39:18Z

0

I would go with:

string.indexOf('33333') > -1

edited Sep 8, 2014 at 16:39

answered Sep 8, 2014 at 0:47

Lucas T.

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