select all options, then exclude the one with that value :
$('#lstCities option[value!="0"]').remove();
Question
I have a select that contain this values:
<select id="lstCities" class="valid" name="City">
<option value="OSNY">OSNY</option>
<option value="dd">dd</option>
<option value="OSffNY">OSffNY</option>
<option value="ANTONY">ANTONY</option>
<option value="0">Autre...</option>
</select>
How can I delete all options but i would like to keep only
<option value="0">Autre...</option>
My problem is my list is dynamic sometimes I have 3,5,7,.... select + the last one <option value="0">Autre...</option>
Solution
select all options, then exclude the one with that value :
$('#lstCities option[value!="0"]').remove();
OTHER TIPS
$('#lstCities option[value!=0]').remove()
You should remove by value and not rely on the position of the option element to keep.
If the option is always the last one, I hope this JavaScript code can help you:
var lastNode = $("#lstCities option").last();
var option = { value:lastNode.val(), text:lastNode.text() };
$('#lstCities').find('option').remove().end().append($('<option>',{
value:option.value,
text:option.text,
}));
Learned from this post
This worked for me:
$("#selectList option").remove();