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Is there any regex (!) which can test if a string contains 3 digits ( never-mind the order) ?

(at least 3 digits. also I be happy to see the exact 3 solution.( if you're kind ))

example :

abk2d5k6 //3 digits
abk25k6d //same here //3 digits

my fail tries :

.+?(?=\d){3}

Thanks.

(only regex solutions please , for learning purpose.).

Is there any regex (!) which can test if a string contains 3 digits ( never-mind the order) ?

(at least 3 digits. also I be happy to see the exact 3 solution.( if you're kind ))

example :

abk2d5k6 //3 digits
abk25k6d //same here //3 digits

my fail tries :

.+?(?=\d){3}

Thanks.

(only regex solutions please , for learning purpose.).

• javascript
• regex

Share Improve this question Follow edited Nov 25, 2012 at 9:28 Royi Namir asked Nov 25, 2012 at 9:14 Royi NamirRoyi Namir 149k144144 gold badges492492 silver badges829829 bronze badges 4

• 2 There is. What have you tried? – John Dvorak Commented Nov 25, 2012 at 9:16
• 1 Also, at most three, at least three or exactly three? – John Dvorak Commented Nov 25, 2012 at 9:16
• @JanDvorak Ive tried a lot. (believe me). I tried with positive look ahead but got stuck with \d+ ( the plus thing. - cause the plus wants it to be sequential). for your second question , lets say at least 3. ( sorry for not clarify it). – Royi Namir Commented Nov 25, 2012 at 9:19
• @RoyiNamir: If you've tried things, show what you've tried, why you thought that was the way to go, and the results you were getting that weren't what you expected. – T.J. Crowder Commented Nov 25, 2012 at 9:24

Add a ment  | 

3 Answers 3

Sorted by: Reset to default 11

Well, for the learning purpose, I suggest to read through this very prehensive tutorial.

Otherwise, the JavaScript regex would probably look something like this:

if you want to make sure that there are at least three digits:

/^(?:\D*\d){3}/

Or if you want to make sure that there are exactly three digits:

/^(?:\D*\d){3}\D*$/

\D is any non-digit character. * allows 0 or more of those. \d is any digit character. {3} repeats the subpattern 3 times. And ^ and $ mark the start and end of the string, respectively.

You could also do something like this:

str.match(/\d/g).length >= 3

This is dead simple, and very clearly shows the intent without a plicated regex.

This isn't tuned for speed though.

Given your ment that you want "at least" three digits, I don't think it needs to be any more plicated than:

/\d.*\d.*\d/

...which is a digit, then zero or more of anything, then another digit, then zero or more of anything, then another digit. As I haven't anchored either end, there's an implicit "zero or more of anything" at both ends.

console.log(!!"abk2d5k6".match(/\d.*\d.*\d/));  // true
console.log(!!"abk25k6d".match(/\d.*\d.*\d/));  // true
console.log(!!"abkd5k6".match(/\d.*\d.*\d/));   // false (I removed a digit)
console.log(!!"abk2k6d".match(/\d.*\d.*\d/));   // false (I removed a digit)

or

/(\d.*){3}/

console.log(!!"abk2d5k6".match(/(\d.*){3}/));  // true
console.log(!!"abk25k6d".match(/(\d.*){3}/));  // true
console.log(!!"abkd5k6".match(/(\d.*){3}/));   // false (I removed a digit)
console.log(!!"abk2k6d".match(/(\d.*){3}/));   // false (I removed a digit)

or at m.buettner points out in the ments below, rather than . you can use \D (not a digit):

/\d\D*\d\D*\d/

or

/(\d\D*){3}/

display(!!"abk2d5k6".match(/\d\D*\d\D*\d/));  // true
display(!!"abk25k6d".match(/\d\D*\d\D*\d/));  // true
display(!!"abkd5k6".match(/\d\D*\d\D*\d/));   // false (I removed a digit)
display(!!"abk2k6d".match(/\d\D*\d\D*\d/));   // false (I removed a digit)

display(!!"abk2d5k6".match(/(\d\D*){3}/));  // true
display(!!"abk25k6d".match(/(\d\D*){3}/));  // true
display(!!"abkd5k6".match(/(\d\D*){3}/));   // false (I removed a digit)
display(!!"abk2k6d".match(/(\d\D*){3}/));   // false (I removed a digit)

Basically, no need for anchors if it's an "at least" match.

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