Question

Here is the code:

$('#date').append(
    '<select id="date">'+
    '<option value="0">- - SELECT - -</option>');
    for(var i in data){
    $('#date').append(
    '<option value="">'+data[i]['date_time']+'</option>');
    });
$('#date').append('</select>');

</select> is always added above for loop. If i replace it with just work select for example, it is appended at the end, where it should be. Why is this happening and how can i fix it?

Answer 1

I believe that jQuery will generate the DOM like this:

<div id="date">
    <select id="date">
        <option value="0">- - SELECT - -</option>
    </select>
    <option value="">foo</option>
    <option value="">bar</option>
    etc...
</div>

Since it is automatically closing the <select> after the first .append(). What you are doing afterwards is appending the options to the <div id="#date"/> rather than the <select> that was appended. I don’t think the final closing </select> will actually do anything either.

If you really want to use append the following JavaScript will add the options to the correct node:

// dummy data object for example
var data = new Array(new Array(), new Array());
data[0].date_time = 2011;
data[1].date_time = 2012;

var options = new Array();
$.each(data, function(index, option) {
    options.push('<option value="' + index + '">'+ option.date_time +'</option>');
});

$('<select id="date"/>').append(options.join('')).appendTo('#date');

Assuming the existing HTML:

<div id="date"></div>

However this does incur an overhead since appending is occurring twice. The faster approach is to build up the options markup as already answered by ShankarSangoli