regular expression to match exactly 5 digits

Question

testing= testing.match(/(\d{5})/g);

I'm reading a full html into variable. From the variable, want to grab out all numbers with the pattern of exactly 5 digits. No need to care of whether before/after this digit having other type of words. Just want to make sure whatever that is 5 digit numbers been grabbed out.

However, when I apply it, it not only pull out number with exactly 5 digit, number with more than 5 digits also retrieved...

I had tried putting ^ in front and $ behind, but it making result come out as null.

Answer 1

This should work:

<script type="text/javascript">
var testing='this is d23553 test 32533\n31203 not 333';
var r = new RegExp(/(?:^|[^\d])(\d{5})(?:$|[^\d])/mg);
var matches = [];
while ((match = r.exec(testing))) matches.push(match[1]);
alert('Found: '+matches.join(', '));
</script>

Answer 2

No need to care of whether before/after this digit having other type of words

To just match the pattern of 5 digits number anywhere in the string, no matter it is separated by space or not, use this regular expression (?<!\d)\d{5}(?!\d).

Sample JavaScript codes:

var regexp = new RegExp(/(?<!\d)\d{5}(?!\d)/g); 
    var matches = yourstring.match(regexp);
    if (matches && matches.length > 0) {
        for (var i = 0, len = matches.length; i < len; i++) {
            // ... ydo something with matches[i] ...
        } 
    }

Here's some quick results.

• abc12345xyz (✓)

• 12345abcd (✓)

• abcd12345 (✓)

• 0000aaaa2 (✖)

• a1234a5 (✖)

• 12345 (✓)

• <space>12345<space>12345 (✓✓)

Answer 3

what is about this? \D(\d{5})\D

This will do on:

f 23 23453 234 2344 2534 hallo33333 "50000"

23453, 33333 50000

Answer 4

My test string for the following:

testing='12345,abc,123,54321,ab15234,123456,52341';

If I understand your question, you'd want ["12345", "54321", "15234", "52341"].

If JS engines supported regexp lookbehinds, you could do:

testing.match(/(?<^|\D)\d{5}(?=\D|$)/g)

Since it doesn't currently, you could:

testing.match(/(?:^|\D)(\d{5})(?=\D|$)/g)

and remove the leading non-digit from appropriate results, or:

pentadigit=/(?:^|\D)(\d{5})(?=\D|$)/g;
result = [];
while (( match = pentadigit.exec(testing) )) {
    result.push(match[1]);
}

Note that for IE, it seems you need to use a RegExp stored in a variable rather than a literal regexp in the while loop, otherwise you'll get an infinite loop.

Answer 5

I am reading a text file and want to use regex below to pull out numbers with exactly 5 digit, ignoring alphabets.

Try this...

var str = 'f 34 545 323 12345 54321 123456',
    matches = str.match(/\b\d{5}\b/g);

console.log(matches); // ["12345", "54321"]

jsFiddle.

The word boundary \b is your friend here.

Update

My regex will get a number like this 12345, but not like a12345. The other answers provide great regexes if you require the latter.