JavaScript substr_count

Returns the number of times a substring occurs in the string

function substr_count (haystack, needle, offset, length) {
    // Returns the number of times a substring occurs in the string  
    // 
    // version: 1109.2015
    // discuss at: http://phpjs.org/functions/substr_count    // +   original by: Kevin van Zonneveld (http://kevin.vanzonneveld.net)
    // +   bugfixed by: Onno Marsman
    // *     example 1: substr_count('Kevin van Zonneveld', 'e');
    // *     returns 1: 3
    // *     example 2: substr_count('Kevin van Zonneveld', 'K', 1);    // *     returns 2: 0
    // *     example 3: substr_count('Kevin van Zonneveld', 'Z', 0, 10);
    // *     returns 3: false
    var pos = 0,
        cnt = 0; 
    haystack += '';
    needle += '';
    if (isNaN(offset)) {
        offset = 0;    }
    if (isNaN(length)) {
        length = 0;
    }
    offset--; 
    while ((offset = haystack.indexOf(needle, offset + 1)) != -1) {
        if (length > 0 && (offset + needle.length) > length) {
            return false;
        } else {            cnt++;
        }
    }
 
    return cnt;}

external links: original PHP docs | raw js source

Examples

» Example 1

Running

1	substr_count('Kevin van Zonneveld', 'e');

Should return

» Example 2

Running

1	substr_count('Kevin van Zonneveld', 'K', 1);

Should return

Dependencies

No dependencies, you can use this function standalone.

Open syntax issues

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Eventually we want all code to pass or at least take into consideration most fixes suggested by JsLint, following this JsLint configuration we’ve decided on.

Authors

Thanks to the following developers, you get to have substr_count goodness in JavaScript.

Kevin van Zonneveld ( original )
Onno Marsman ( bugfixed )

Comments

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Use:

[CODE]
your_stuff('here');
[/CODE]

for proper code formatting
By submitting code here you are allowing us to use it in php.js hence dual licensing it under the MIT and GPL licenses

Rafal
6 Oct '11

Great!

Świetne, bardzo mi się przydało.
Greetings from Poland - Rafal

Robert Eisele
5 Sep '11

What about this optimization?

function substr_count(haystack, needle, offset, length) {

    haystack+= "";
    needle+= "";

    if (isNaN(offset)) {
        offset = 0;
    }
    if (isNaN(length)) {
        length = haystack.length - offset;
    }

    haystack = haystack.substr(offset, length);

    return (haystack.length - haystack.replace(new RegExp(needle, 'g'), "").length) / needle.length;
}

Kevin van Zonneveld
24 Jul '09

@ Shock: I think I would have to make the same point as here:
http://phpjs.org/functions/substr_count:559#comment_626

Don't you agree?

Shock
20 Jul '09

// Maybe something like that?

function substr_count( haystack, needle, offset, length ) {
    if (isNaN(offset)) {
        offset = 0;
    }
    if (isNaN(length)) {
        length = haystack.length - offset;
    }
    if (length <= 0 || (length + offset) > haystack) {
        return false
    } else {
        return haystack.substr(offset,length).split(needle).length - 1;
    }
}

sorry for doublePost

Shock
20 Jul '09

// Maybe something like that?
[CODE]function substr_count( haystack, needle, offset, length ) {
if (isNaN(offset)) {
offset = 0;
}
if (isNaN(length)) {
length = haystack.length - offset;
}
if (length <= 0 || (length + offset) > haystack) {
return false
} else {
return haystack.substr(offset,length).split(needle).length - 1;
}
}[/CODE

Kevin van Zonneveld
25 Jan '09

Well we could benchmark it of course. But I think that when your doing this on really large strings, you also end up with really large arrays, that you really don't need. So my guess is that your approach will be more memory intensive don't you agree?

antimatter15
18 Jan '09

Wouldn't it be faster to just use split().length-1?

[CODE="Javascript"]function substr_count( haystack, needle, offset, length ) {
if(!isNaN(offset)){
if(!isNaN(length)){
haystack=haystack.substr(offset,length);
}else haystack = haystack.substr(offset)
}
haystack = haystack.split(needle).length-1
return haystack<0?false:haystack;
}[/CODE]

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