regex - JavaScript: add or subtract from number in string - Stack Overflow

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I have a string that looks like "(3) New stuff" where 3 can be any number.
I would like to add or subtract to this number.

I figured out the following way:

var thenumber = string.match((/\d+/));
thenumber++;
string = string.replace(/\(\d+\)/ ,'('+ thenumber +')');

Is there a more elegant way to do it?

I have a string that looks like "(3) New stuff" where 3 can be any number.
I would like to add or subtract to this number.

I figured out the following way:

var thenumber = string.match((/\d+/));
thenumber++;
string = string.replace(/\(\d+\)/ ,'('+ thenumber +')');

Is there a more elegant way to do it?

• javascript
• regex

Share Improve this question Follow edited May 1, 2010 at 13:00 Peter Mortensen 31.6k2222 gold badges110110 silver badges133133 bronze badges asked Feb 2, 2009 at 18:14 yoavfyoavf 21.3k99 gold badges3838 silver badges3838 bronze badges

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5 Answers 5

Sorted by: Reset to default 5

Another way:

string = string.replace(/\((\d+)\)/ , function($0, $1) { return "(" + (parseInt($1, 10) + 1) + ")"; });

I believe Gumbo was on the right track

"(42) plus (1)".replace(/\((\d+)\)/g, function(a,n){ return "("+ (+n+1) +")"; });

Short of extending the String object, it looks good to me.

String.prototype.incrementNumber = function () {
  var thenumber = string.match((/\d+/));
  thenumber++;
  return this.replace(/\(\d+\)/ ,'('+ thenumber +')');
}

Usage would then be:

alert("(2) New Stuff".incrementNumber());

I believe your method is the best elegant you can have for following reasons:

• since the input is not a "clean" number, you do need to involve some sort of string parser. Using regular expressions is the very code-efficient method to do it
• by looking at the code, it's clear what it does

short of wrapping this into a function, I don't think there's much more to be done

As galets says, I don't think your solution is a bad one but here is a function that will add a specified value to a number in a specified position in a string.

var str = "fluff (3) stringy 9 and 14 other things";

function stringIncrement( str, inc, start ) {
    start = start || 0;
    var count = 0;
    return str.replace( /(\d+)/g, function() {
        if( count++ == start ) {
            return(
                arguments[0]
                .substr( RegExp.lastIndex )
                .replace( /\d+/, parseInt(arguments[1])+inc )
            );
        } else {
            return arguments[0];
        }
    })
}

// fluff (6) stringy 9 and 14 other things :: 3 is added to the first number
alert( stringIncrement(str, 3, 0) );

// fluff (3) stringy 6 and 14 other things :: -3 is added to the second number
alert( stringIncrement(str, -3, 1) );

// fluff (3) stringy 9 and 24 other things :: 10 is added to the third number
alert( stringIncrement(str, 10, 2) );

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