Array of digit as string to simple int conversion problem

Question

Console debug shows me that array is ex. ["2"], but I need [2].

Why casting doesnt'work?

function filterElements(deals) {

    var result = deals,
        categories= $('#deals-categories').data('current_category');
        if (categories != undefined && categories.length > 0) {
            for (var i; i < categories.length; i++) {
                categories[i] = parseInt(categories[i]);
            }
            console.log(categories, 'cats');
                result = $.grep(result, function(e) {
                    return $.inArray(e.category_id, categories) != -1;
     });                
        }
    return result;
}

Matt Ball · Accepted Answer · 2011-08-29 13:10:11Z

4

You need to initialize var i = 0 in the loop declaration.

Full code cleanup:

function filterElements(deals) {

    var result = deals,
        categories = $('#deals-categories').data('current_category');

        if (categories && categories.length) {
            for (var i=0; i<categories.length; i++) {
                categories[i] = parseInt(categories[i], 10);
            }
            console.log(categories, 'cats');
            result = $.grep(result, function(e) {
                return $.inArray(e.category_id, categories) !== -1;
            });                
        }

    return result;
}

answered Aug 29, 2011 at 13:10

Matt Ball

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1 Comment

Matt Ball Over a year ago

The only weirdness I see now is using $.inArray() inside of $.grep().

Michael Sagalovich · Accepted Answer · 2011-08-29 13:16:47Z

0

use categories[i] * 1 to cast

parseInt works in a bit unexpected way sometimes :)

parseInt("010") will return 8 in some browsers, and 10 in others: http://www.w3schools.com/jsref/jsref_parseInt.asp

edited Aug 29, 2011 at 13:16

answered Aug 29, 2011 at 13:09

Michael Sagalovich

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5 Comments

Michael Sagalovich Over a year ago

I think Matt Ball explains the issue, while parseInt is still not recommended when dealing with strings which are for sure ints.

Matt Ball Over a year ago

@Michael why not? Ohhh, w3schools... /facepalm. This is why JSLint recommends that you always specify the radix in parseInt calls (until ECMAScript 5). A much better resource than w3schools: developer.mozilla.org/en/JavaScript/Reference/Global_Objects/…

Michael Sagalovich Over a year ago

yeah, radix solves the issue, but requires too much letters :)

Matt Ball Over a year ago

The better alternatives to parseInt() are Number or unary +, not *1.

Matt Ball Over a year ago

See stackoverflow.com/questions/4090518/… and stackoverflow.com/questions/850341/… and stackoverflow.com/questions/594625/parseint-alternative

Alex Text · Accepted Answer · 2011-08-29 13:39:42Z

0

Are you sure? This is an example similar to yours:

var strings = ["1", "2", "3"];
var valueAsInt = 0;

for(var i = 0; i < strings.length; i++){
   valueAsInt = parseInt(strings[i]);
   if(typeof(valueAsInt) == 'number'){
      alert('Is an integer');
   }
}

The message 'Is an integer' is shown three times. I think in your code the parser works, but maybe later, the value is converted to string by comparing with other string or, maybe some concatenation.

answered Aug 29, 2011 at 13:39

Alex Text

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