使用正则表达式仅检索7位数的字符串

Question

In short, the page contains about 10 divs, each containing letters and numbers and I'm trying to return all occurrences of exactly 7 digits, but can have other, non-digit characters before and after.

eg.

"q1234567" should return 1234567

"q1234567q" should return 1234567

"q1234567q1234567q12345678q" should return 1234567 and 1234567

"12345678" should NOT be returned

To be more specific, an example of an entire string:

q1234567q
q1234567q
q12345678q
q1234567q123456789q123456q1324567q1234567
1234567
1
12
123
1234
12345
q12345q
q1234
12345q
123

I tried doing this via regex and got as far as

/\d{7}(?=\D|$)/g

but JavaScript doesn't play well with the lookbehind.. How can I get around this without involving a whole new library?

Answer 1

This regex should work:

/^\D*\d{7}\D*$/

Online Demo: http://regex101.com/r/nE5eI6

UPDATE: As per edited question and comments below you can use this regex:

(?:^|\D)(\d{7})(?=\D|$)

And use matched group #1 for your output.

Demo: http://regex101.com/r/wL4oW1

Answer 2

You could maybe use something like this?

var regex = /(?:^|\D)(\d{7})(?!\d)/g;
var s = "q1234567q123456789q123456q1324567q1234567";
var match, matches=[];

while ( (match=regex.exec(s)) !== null ) {
    matches.push(match[1]);
}

alert(matches);

jsfiddle demo

Answer 3

怎么样: (?<=\\D)[0-9]{7}(?=\\D|$)|^[0-9]{7}(?=\\D|$) ？

Answer 4

var regex = /(\b|\D)(\d{7})(\b|\D)/g
function get7digits(s) {
    var md, matches=[];
    while( (md=regex.exec(s)) !== null ) matches.push(md[2]);
    return matches;
}

get7digits('q1234567q123456789q123456q1324567q1234567')

["1234567", "1324567"]

Answer 5

How about this one:

/(^|\\D)\\d{7}(\\$|\\D)/gm

http://regex101.com/r/iT5cR3