I was using javascript to detect for specific key strokes and while writing the method I thought I'd try regular expressions and the test() method and came up with:

if (/8|9|37|38|46|47|48|49|50|51|52|53|54|55|56|57|96|97|98|99|100|101|102|103|104|105|110/.test(num)) {
    // do something if there's a match
}

This doesn't seem to work 100% as some values seem to make it past the regex test, such as 83. I've since moved on, but I'm still curious as to why this didn't work.

I was using javascript to detect for specific key strokes and while writing the method I thought I'd try regular expressions and the test() method and came up with:

if (/8|9|37|38|46|47|48|49|50|51|52|53|54|55|56|57|96|97|98|99|100|101|102|103|104|105|110/.test(num)) {
    // do something if there's a match
}

This doesn't seem to work 100% as some values seem to make it past the regex test, such as 83. I've since moved on, but I'm still curious as to why this didn't work.

• javascript
• regex

• you say you are detecting key strokes, when it passed 83, are you sure it didn't really pass the 8 condition? – house9 Commented May 7, 2010 at 20:08
• That is the UGLIEST regex ever. Gah – Mitch Dempsey Commented May 7, 2010 at 20:08
• I'm thinking that you probably have a bigger design problem than the actual methodology of this little if statement. Why do you have such a large set of numbers to match against in the first place? Is that a whitelist of keystrokes? – eyelidlessness Commented May 7, 2010 at 20:09
• @house9, the regex as it exists would definitely match 83. – eyelidlessness Commented May 7, 2010 at 20:11
• 2 @webdestroya, it's far from the ugliest regex ever. In fact, it's one of the cleanest I've seen, if a bit misguided. – eyelidlessness Commented May 7, 2010 at 20:12

4 Answers 4

This is the pletely wrong way to do it.

To answer the question, the regex is matching part of your string. The string 83 passes by matching the 8.
You need to anchor your regex by putting ^( at the beginning and )$ at the end.

The correct way to do this is to make an array of valid numbers, and pare using parseInt.

For example:

var validNumbers = [ 8, 9, 37, 38, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 110 ];

if (validNumbers.indexOf(parseInt(num, 10)) >=0 ) {
    //Match
}

You'll need an indexOf function for IE:

if (!Array.prototype.indexOf) {
    Array.prototype.indexOf = function(needle) {
        for(var i = 0; i < this.length; i++) {
            if(this[i] === needle) {
                return i;
            }
        }
        return -1;
    };
}

You need to specify the start and end of the string. Otherwise 8 in 8|… will match the 8 in 83:

/^(8|9|37|38|46|47|48|49|50|51|52|53|54|55|56|57|96|97|98|99|100|101|102|103|104|105|110)$/.test(num)

But you should rather use numeric parison. If you don’t like to list every number, you can use ranges like this:

function foo(number, numbers) {
    for (var i=0; i<numbers.length; ++i) {
        if (numbers[i] === number) {
            return true;
        } else if (numbers[i].constructor === Array) {
            if (numbers[i][0] <= number && number <= numbers[i][1]) {
                return true;
            }
        }
    }
    return false;
}
var numbers = [8, 9, 37, 38, [46, 57], [96, 105], 110];
if (foo(num, numbers)) {
    // …
}

If you make a regular expresion like /\b(100|101)/g it will match only 100 and 101 and not 5100, 5101 or ...101...;

The only problem with this is if your are using negative numbers, e.g in case of 101 and -101 both match with the regexp.

I know this because is what I'm facing and want to avoid.

I can share with you an example:

let  array = [89, 7, 92, 78, 899, 65, 56, 92, 922, 292, 289, 389, 2879, 2932, 8999];

• I want to find how many instances of the number 92 exist in that array. Therefore I am searching for the precise number of 92. And I want to use Regular Expressions in Javascript.

• First part es with the transformation of the array into a string:

  let strCheck = array.toString(); // "89,7,92,78,899,65,56,92,922,292,289,389,2879,2932,8999"

let regex = /\b92\b/g;

I used the flag g for global so that it finds all the matches and the \b word boundary as described in MDN.

let arrResult = strCheck.match(regex); // ["92", "92"]

That's it. The tricky part was first to acknowledge that Regular Expressions work with strings, the second was that once I got the string I had to think about getting the number I wanted not as a number but as a string which was going to be surrounded by other characters and reaching out to those other characters, helped to find the solution.