循环时如何忽略禁用的复选框值

Question

below is my code:

Cell.setAttribute('type', 'checkbox');
Cell.setAttribute('name', "Size");

based on condition I will set some values to 'checked' and 'disabled'.

Cell.setAttribute('checked', 'true');
Cell.setAttribute('disabled', 'true');   

Here 'Size' contains some values which are disabled, both unchecked and checked.

So whenever I am trying to retrieve the checked values, the default disabled checked boxes are also being returned.

I just want the checked values; not the disabled checked values in SelectedDetails method:

var testCaseArray =[]; 
var checkbox = document.getElementsByName('Size');

for(var i=0; i<checkbox.length; i++){
    if(checkbox[i].checked){ 
        testCaseArray.push(checkbox[i].value);
    }
}

return testCaseArray;

Answer 1

Retrieve all checked checkbox elements in a series, except the disabled one(s), into an Array :

var chkd = [].slice.call(document.querySelectorAll('input:checked'))
            .filter(function(e) {
               return null == e.getAttribute('disabled');
             });

You can also use this selector to retrieve a Array-like list with the same selection:

document.querySelectorAll('input:checked:not(disabled)')

jsFiddle

Answer 2

You need to add in the check for disabled, as well as the check for it being checked

var testCaseArray =[]; 
var checkbox = document.getElementsByName('Size');

for(var i=0; i<checkbox.length; i++){
    if(checkbox[i].disabled){
        continue;
    }
    if(checkbox[i].checked){ 
        testCaseArray.push(checkbox[i].value);
    }
}

return testCaseArray;

Answer 3

Since JavaScript has short-circuit evaluation , just add the condition to your if statement:

var testCaseArray = [],
    checkbox = document.getElementsByName('Size');

for (var i = 0; i < checkbox.length; ++i) {
    if (!checkbox[i].disabled && checkbox[i].checked) {
        testCaseArray.push(checkbox[i].value);
    }
}

return testCaseArray;

The JavaScript engine will automatically stop evaluating if !checkbox[i].disabled returns false.