Javascript function returning false even object found

Question

I have a Product javascript array, which contains all the products information. Created a function which iterate on this array and find the product by matching id.

var products = JSON.parse('[{"Product":{"id":"1","name":"My Product","description":"This is my new product","price":"10.00","currency":"$","stock":"0","image":"/image1.png"}},{"Product":{"id":"5","name":"Dummy Product 2","description":"Some dummy text goes here.","price":"10.00","currency":"$","stock":"100","image":"image2.jpg"}}]');

$(document).ready(function(){
    console.log(products);
    alert(findProduct(5)); //it will returns false everytime, evan it has matching product
});

function findProduct(product_id){
    $.each(products, function(k, v){
        if(v.Product.id == product_id){
            console.log(v);
            return products[k]; //or return 'v'
        }
    });
    return false;
}

Check this Demo

Function returns false each time, even though it found the matching product id, don't know why? If I store the matching array key in a variable and after iteration, return the value of that key it returns proper object. But that's not proper way, cause I want to stop iteration and return the value if object found.

Arun P Johny · Accepted Answer · 2014-02-05 09:33:09Z

1

You are always returning false from findProduct, if the item is found you are returning from the $.each() callback method, but that is not reflected in the value returned by findProduct method.

function findProduct(product_id) {
    var found = false;
    $.each(products, function (k, v) {
        if (v.Product.id == product_id) {
            console.log(v);
            found = products[k];
            //return false to prevent further iteration
            return false;
        }
    });
    return found;
}

answered Feb 5, 2014 at 9:33

Arun P Johny

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Comments

Minko Gechev · Accepted Answer · 2014-02-05 09:32:16Z

1

Use this instead:

function findProduct(product_id){
    var result = false;
    $.each(products, function(k, v){
        if(v.Product.id == product_id){
            console.log(v);
            result = products[k]; //or return 'v'
            return;
        }
    });
    return result;
}

And here is the problem:

function findProduct(product_id){
    $.each(products, function(k, v){
        if(v.Product.id == product_id){
            console.log(v);
            return products[k]; //returns from the $.each callback
        }
    });
    return false;
}

answered Feb 5, 2014 at 9:32

Minko Gechev

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2 Comments

Tony Over a year ago

return result; maybe?

Minko Gechev Over a year ago

Yep, fixed it. Thanks for the note.

Ivan Seidel · Accepted Answer · 2014-02-05 09:36:38Z

0

No need for jQuery's each function. Use libraries only when they are needed.

The problem is, that the 'each' method, is in fact ANOTHER function. Returning from it means nothing to the 'findProduct' function.

function findProduct(product_id){
    for(var k in products){
        var product = products[k];
        if(product && product.Product && product.Product.id == product_id)
            return product;
    }
    return false;
}

answered Feb 5, 2014 at 9:36

Ivan Seidel

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