Regular expression to require 10 digits without co

问题:

I need a regex to limit number digits to 10 even if there are spaces.

For example it allows 06 15 20 47 23 the same as 0615204723.

I tried: ^\d{10}$, but how do I ignore spaces?

Also, the number should start with 06 or 07. (Editor's note, the last requirement was from comments by the OP.)

回答1:

This will do it in most regex systems. (You need to state what language you are using!)

/^\s*(0\s*[67]\s*(?:\d\s*){8})$/

But assuming you really only want the number, then go ahead and break it into 2 steps.

For example, in JavaScript:

var s = ' 06 15 20 47   34 ';
s = s.replace (/\s/g, "");
isValid = /^0[67]\d{8}$/.test (s);

回答2:

You can do:

"^(\d *){10}$"

回答3:

I had the same problem, I noticed if I grouped each digit with any number of spaces padding the front and back, I got the solution I was looking for. In this example, ignores all spaces

^(\s*\d\s*){10}$

eg. "1234567890"
    " 12  34 567   89 0    "

In a more complicated example, this ignored all spaces, dashes, round braces and spaces.

^([\s\(\)\-]*\d[\s\(\)\-]*){10}$

eg. "[phone]"
    "   (   12 (3)( - 45 - 6-78 )(90 (((--  "

Test them out here, it will provide you with the code for your given system: http://www.myregextester.com/index.php

回答4:

Slightly preferable:

"^\s?(\d\s?){10}$"

Where \s is any whitespace and \d is a digit. This allows for leading as well as infixed and trailing spaces.

回答5:

(\s*\d\s*){0,10}

This should allow leading and trailing spaces around digits, and match 0 to 10 digits.

回答6:

var str = " a b c d e f g "; 
var newStr = str.replace(/\s+/g, '');