Count number of matches of a regex in Javascript-ThrowExceptions

Exception or error:

I wanted to write a regex to count the number of spaces/tabs/newline in a chunk of text. So I naively wrote the following:-

numSpaces : function(text) { 
    return text.match(/\s/).length; 
}

For some unknown reasons it always returns 1. What is the problem with the above statement? I have since solved the problem with the following:-

numSpaces : function(text) { 
    return (text.split(/\s/).length -1); 
}
How to solve:

tl;dr: Generic Pattern Counter

// THIS IS WHAT YOU NEED
const count = (str) => {
  const re = /YOUR_PATTERN_HERE/g
  return ((str || '').match(re) || []).length
}

For those that arrived here looking for a generic way to count the number of occurrences of a regex pattern in a string, and don’t want it to fail if there are zero occurrences, this code is what you need. Here’s a demonstration:

/*
 *  Example
 */

const count = (str) => {
  const re = /[a-z]{3}/g
  return ((str || '').match(re) || []).length
}

const str1 = 'abc, def, ghi'
const str2 = 'ABC, DEF, GHI'

console.log(`'${str1}' has ${count(str1)} occurrences of pattern '/[a-z]{3}/g'`)
console.log(`'${str2}' has ${count(str2)} occurrences of pattern '/[a-z]{3}/g'`)

Original Answer

The problem with your initial code is that you are missing the global identifier:

>>> 'hi there how are you'.match(/\s/g).length;
4

Without the g part of the regex it will only match the first occurrence and stop there.

Also note that your regex will count successive spaces twice:

>>> 'hi  there'.match(/\s/g).length;
2

If that is not desirable, you could do this:

>>> 'hi  there'.match(/\s+/g).length;
1

###

As mentioned in my earlier answer, you can use RegExp.exec() to iterate over all matches and count each occurrence; the advantage is limited to memory only, because on the whole it’s about 20% slower than using String.match().

var re = /\s/g,
count = 0;

while (re.exec(text) !== null) {
    ++count;
}

return count;

###

(('a a a').match(/b/g) || []).length; // 0
(('a a a').match(/a/g) || []).length; // 3

Based on https://stackoverflow.com/a/48195124/16777 but fixed to actually work in zero-results case.

###

('my string'.match(/\s/g) || []).length;

###

This is certainly something that has a lot of traps. I was working with Paolo Bergantino’s answer, and realising that even that has some limitations. I found working with string representations of dates a good place to quickly find some of the main problems. Start with an input string like this:
'12-2-2019 5:1:48.670'

and set up Paolo’s function like this:

function count(re, str) {
    if (typeof re !== "string") {
        return 0;
    }
    re = (re === '.') ? ('\\' + re) : re;
    var cre = new RegExp(re, 'g');
    return ((str || '').match(cre) || []).length;
}

I wanted the regular expression to be passed in, so that the function is more reusable, secondly, I wanted the parameter to be a string, so that the client doesn’t have to make the regex, but simply match on the string, like a standard string utility class method.

Now, here you can see that I’m dealing with issues with the input. With the following:

if (typeof re !== "string") {
    return 0;
}

I am ensuring that the input isn’t anything like the literal 0, false, undefined, or null, none of which are strings. Since these literals are not in the input string, there should be no matches, but it should match '0', which is a string.

With the following:

re = (re === '.') ? ('\\' + re) : re;

I am dealing with the fact that the RegExp constructor will (I think, wrongly) interpret the string '.' as the all character matcher \.\

Finally, because I am using the RegExp constructor, I need to give it the global 'g' flag so that it counts all matches, not just the first one, similar to the suggestions in other posts.

I realise that this is an extremely late answer, but it might be helpful to someone stumbling along here. BTW here’s the TypeScript version:

function count(re: string, str: string): number {
    if (typeof re !== 'string') {
        return 0;
    }
    re = (re === '.') ? ('\\' + re) : re;
    const cre = new RegExp(re, 'g');    
    return ((str || '').match(cre) || []).length;
}

###

how about like this

function isint(str){
    if(str.match(/\d/g).length==str.length){
        return true;
    }
    else {
         return false
    }
}

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