Check field values against a preset object

Question

I want to check some fields values in a form against an existing object's values. If the values match, change a class value (it will add a sign). The following code is working, but I would like to know if it can be improved.

Preset obj:

var obj = {
 preset_a:{
 input:{
 field_a:"value",
 ...
 field_b:"value",
 },
 select:{
 ...
 },
 radio:{
 ...
 }
 },
 preset_b:{
 input:{
 field_a:"value",
 ...
 field_b:"value",
 },
 select:{
 ...
 },
 radio:{
 ...
 }
 }
};

The script:

$(document).ready(function(){ 
 for (var preset in obj){
 $.each(obj[preset], function(type){
 var flag = true;
 $.each(obj[preset][type], function(k, v){ 
 if (type == 'radio'){
 flag = ((v) == $('#' + k + ' > input[type="radio"]:checked').val());
 } else {
 flag = ((v) == $(type + '#' + k).val());
 } 
 if (flag == false){ 
 return false;
 } 
 });
 if (flag == false){
 $(preset + ' i#id').attr('class', 'radio-unchecked');
 return false;
 } else {
 $(preset + ' i#id').attr('class', 'radio-checked on');
 }
 });
 }
});

Answer 1

Inside one of the $.each calls, you could simplify this part:

if (flag == false){ 
 return false;
}

to this:

return flag;

When the value is true, $.each will continue normally, and stop after it is false, so the behavior remains the same.

---

Instead of:

if (flag == false) {

It's more common to write:

if (! flag) {

For simplicity, it might be a good idea to flip the condition:

if (flag) {
 $(preset + ' i#id').attr('class', 'radio-checked on');
} else {
 $(preset + ' i#id').attr('class', 'radio-unchecked');
 return false;
}

And to reduce code duplication, I would extract the dom lookup that is common in both branches of the if-else:

var preset = $(preset + ' i#id');
if (flag) {
 preset.attr('class', 'radio-checked on');
} else {
 preset.attr('class', 'radio-unchecked');
 return false;
}

---

The two innermost loops use $.each, but the outer loop is a traditional for. For the sake of consistency, it might be good to use $.each for all.

• n.h.: Thank you very very much for your tips and explanations. Could you be so kind to take a look [codereview.stackexchange.com/questions/95959/… that is more or less related.