What's wrong with my regular expression for IP addresses?

Question

I'm trying to check strings to see it they are a valid IP address:

export function checkIpIsValid(ip) {                                          
    const regex = new RegExp("^([1-2]?[0-9]{1,2}\.){3}[1-2]?[0-9]{1,2}$")     
    return regex.test(ip);                                                    
}

But this test is failing because it return true:

expect(checkIpIsValid("1.1.1.1")).toBe(true);            
expect(checkIpIsValid("152.154.22.35")).toBe(true);      
expect(checkIpIsValid("552.154.22.35")).toBe(false);

// These three are fine
// But the one below is not, it doesn't contain 4 parts so it should return false. Instead, it is returning true.      

expect(checkIpIsValid("154.22.35")).toBe(false);     


// Expected value to be (using ===):
//   false
// Received:
//   true

The thing is, when I check it on https://regex101.com/ it works fine on that case...

No, that does not match. The regex is not correct for IPs, but it does not match the input. Compare. https://regex101.com/r/t4Ooc8/1 — Tomalak, Dec 03 '17 at 10:32
@tomalak Yep, in regex101.com that appears to be correct, but when you run in on javascript, i'm getting a match with the exact same regular expression. — Enuff, Dec 03 '17 at 10:51

score 4 · Accepted Answer · answered Dec 03 '17 at 10:57

Don't use new RegExp(). This expects a string input, but your expression is not escaped for string. You just enclosed it in double quotes, that doesn't work.

Either use a regex literal:

function checkIpIsValid(ip) {
    const regex = /^([1-2]?[0-9]{1,2}\.){3}[1-2]?[0-9]{1,2}$/;    
    return regex.test(ip);                                                    
}

or escape he expression properly:

function checkIpIsValid(ip) {
    const regex = new RegExp("^([1-2]?[0-9]{1,2}\\.){3}[1-2]?[0-9]{1,2}$");    
    return regex.test(ip);                                                    
}

JS has regex literals exactly because that avoids the need for double-escaping. Use new RegExp() only for dynamically built expressions, not for fixed ones.

score 1 · Answer 2 · answered Dec 03 '17 at 10:55

1

function checkIpIsValid(ip) {
    return /^(?:\d{1,3}\.){3}\d{1,3}$/.test(ip);
}

checkIpIsValid('123.12.12'); // returns false
checkIpIsValid('123.12.12.12'); // returns true

answered Dec 03 '17 at 10:55

Prashant

7,340
2
25
24

Humm, the thing is that one will match my third test case: `552.154.22.35` and it shouldn't – Enuff Dec 03 '17 at 10:57

What's wrong with my regular expression for IP addresses?

2 Answers2