javascript regex for ip address with asterisk masking

Question

I would like to make a regular expression for an IP address with asterisk(*) which matches these below:

The digit 127.0 could be any number between 0 to 255.

**[TRUE]**
127.*.*.*
127.0.*.*
127.0.0.*

**[FALSE]**
127.*.*.1
127.*.0.1
127.0.*.1

What I have made until now is...and of course, failed to make it out. I totally got lost..

_regex = function(value) {
    var _match = /^(?:(\d|1\d\d|2[0-4]\d|25[0-5]))\.(?:(\*|\d|1\d\d|2[0-4]\d|25[0-5]))\.(\*|(?:\d{1,2}(?:.\d{1,3}?)))\.(\*|(?:\d{1,3}(?:\*?)))$
    if(_match.test(value)){
        //do something;
    }
}

If you give me any chance to learn this, would be highly appreciated. Thanks.

@ CertainPerformance @ÁlvaroTouzón Sorry for my mistake. I didn't copy my script and rushed to write this, so made many mistakes.. — Inye Yu
– Inye Yu, Commented Jan 8, 2019 at 11:52

BenG · Accepted Answer · 2019-01-08 12:12:31Z

2

I think what you are looking for is a negative look ahead to make sure no number follows an asterisk.

Like so: (\*(?!.*\d))

working example:

var ips = [
  '127.*.*.*',
  '127.0.*.*',
  '127.0.0.*',
  '127.*.*.1',
  '127.*.0.1',
  '127.0.*.1'
];

var regex = /^((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?|(\*(?!.*\d)))(\.|$)){4}$/;

for(var i = 0; i < ips.length; i++){
  console.log(ips[i] + ': ' + regex.test(ips[i]));
}

answered Jan 8, 2019 at 12:12

BenG

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