select option remove() method on removing multiple options

Question

I'm having a select box which contains 17 values/options.

<select id="_class" name="_class">
    <option value="0"></option>
    <option value="1">PREKG</option>
    <option value="2">LKG</option>
    <option value="3">UKG</option>
    <option value="4">I</option>
    <option value="5">II</option>
    <option value="6">III</option>
    <option value="7">IV</option>
    <option value="8">V</option>
    <option value="9">VI</option>
    <option value="10">VII</option>
    <option value="11">VIII</option>
    <option value="12">IX</option>
    <option value="13">X</option>
    <option value="14">XI</option>
    <option value="15">XII</option>
    <option value="16">XIII</option>                              
</select>

Now, I'm trying to remove the option from 5 to 8. For that, I'm using the following JavaScript code.

<script type="text/javascript">
 var _class = document.getElementById("_class");
 for(var i=1; i < _class.length;i++) {
   if(i>=5 && i<=8) {
     _class.remove(i);
   }
 }
</script>

But, I'm not getting the expected result, because every time as if the for loop runs, the order of the option is getting changed.

How can I get the desired result?

Here I've attached the Fiddle.

Answer 1

By iterating through the options incrementally, the indices of the options after an option that is removed will be reduced by one as soon as the option is removed (ie: removing the option at index 5 causes options 6 and above to now be indexed as options 5 and above).

To cater for this, reverse the order of the loop to count downwards:

var _class = document.getElementById("_class");

for(var i=Math.min(_class.options.length, 8); i >= 5; i--) {
  _class.remove(i);
}

http://jsfiddle.net/Gy8j8/2/