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Can anyone send javascript code to validate the network mac address (eg. 02:41:6d:22:12:f1) It accepts value from 00:00:00:00:00:00 to ff:ff:ff:ff:ff:ff. On keypress event of textbox, I need to allow characters 0-9, a-f and : (colon). What I have so far is

macPattern = /^([0-9a-fA-F]{2}[:-]){5}[0-9a-fA-F]{2}$/i;

With this I am able throw exception for ff:ff:ff:ff:ff:ff but I also need to throw an exception for 00:00:00:00:00:00. My pattern is not throwing an exception.

Could you please give me a pattern through which I should able to throw an exception for both ff:ff:ff:ff:ff:ff and 00:00:00:00:00:00.

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    Rather than trying to make a more complex regex, why not just first check the input for the values ff:ff:ff:ff:ff:ff and 00:00:00:00:00:00, and then if it is neither of those, run it through the regex? Commented Aug 20, 2013 at 13:12
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    I would just use the regex as is, and add a quick check for the two known exceptions (just using !=). Commented Aug 20, 2013 at 13:13
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    A regular expression is not synonymous with "do all the logic needed to validate input." If you have a requirement like excluding ff:ff:ff:ff:ff:ff and 00:00:00:00:00:00, it would be much better to exclude those cases using ordinary programming logic as @ajp15243 suggested twice, rather than "could you please give me compltely pattern." P.S. Is someone telling you that you have to do it all in a single regular expression? If so, who and why? Is this homework? This definitely doesn't seem to meet any real-world standard for ensuring input is an actual MAC address. Commented Aug 20, 2013 at 13:26
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    stackoverflow.com/questions/4260467/… Commented Aug 20, 2013 at 13:30
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    After some checking, I see that there are 281,474,976,710,656 possible valid MAC addresses. That is exactly 2^48, which means that both of the cases you are trying to exclude are theoretically possible. Commented Aug 20, 2013 at 13:36

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var MACAddress = document.getElementById("MACAddress"); var MACRegex=new RegExp("^([0-9a-fA-F][0-9a-fA-F]:){5}([0-9a-fA-F][0-9a-fA-F])$"); MACRegex.test(MACAddress);
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var re = /^(?!(?:ff:ff:ff:ff:ff:ff|00:00:00:00:00:00))(?:[\da-f]{2}:){5}[\da-f]{2}$/i; // ^------------blacklist------------^ ^----------pattern---------^ re.test('ff:ff:ff:ff:ff:ff'); // false re.test('ff:ff:ff:ff:ff:fe'); // true re.test('00:00:00:00:00:00'); // false re.test('00:00:00:00:00:01'); // true // and of course re.test('00:00:00:00:01'); // false re.test('00:00:00:00:00:0g') // false
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Oh if we are answering about validating mac, here is mine: but this still doesn't answer the question: how to prevent characters that don't make up a mac address.

function isValidMac(mac) { var a = mac.split(':'); if (a.length !== 6) { return false; } for (var i=0; i<6; i++) { var s = "0x"+a[i]; if (s>>0 === 0 || s.length != 4) { return false; } } return true; }
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I think that while the other answers are all valid, they missed the keypress aspect of the OP's question. While that might not be important in this instance, I believe that the UX can be improved.

I would suggest;

-validating length =12 -accepting {0-9,a-f,A-F}, -alert {g-z,G-Z) (invalid character) -ignore all others (including Tab, cr, lf, crlf) -confirm exit after the 12th character -display 3 forms; raw, couplet, quad

I have not yet had a chance to code but will submit and amend on completion

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My answer is:

^([0-9a-fA-F]{2}[:-]){5}[0-9a-fA-F][1-9a-eA-E]$
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Why must only the last digit not be 0 or f? What about any of the other digits? That regex ends up not matching on any MAC address that ends in 0 or f (12:34:56:78:9a:bf, for instance, would not be matched), rather than excluding the case of all 0s or all fs.

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