Reputation: 2442
I have the following dropdown slect:
<select id="start" onchange="calcRoute();"></select>
I need to populate this with values I get from a Javascrippt array of this type:
{"name":"Hotel1","address":"Bondi Beach","lat":"6.369869","lng":"80.042055","thumb_path":"img\/hotel1.jpg"},{"name":"Hotel2","address":"Coogee Beach","lat":"6.025822","lng":"80.305727","thumb_path":"img\/hotel2.jpg"},{"name":"Hotel3","address":"Cronulla Beach","lat":"6.691994","lng":"79.928771","thumb_path":"img\/hotel3.jpg"},{"name":"Hotel4","address":"Manly Beach","lat":"6.887130","lng":"80.093420","thumb_path":"img\/hotel4.jpg"},{"name":"Hotel5","address":"Maroubra Beach","lat":"7.362609","lng":"79.825005","thumb_path":"img\/hotel5.jpg"}
I tried the following, but it doesnt work.
for (i = 0; i < locations.length; i++) {
$('#start select').append('<option value=' + locations[i]['name'] + '>' + locations[i]['name'] + '</option>');
alert("sdgsfg");
}
Please suggest a fix.
Upvotes: 1
Views: 1273
Reputation: 193261
The selector $('#start select')
will select the <select>
element inside the #start
element which not work as the id of the select
itself is start
. To select the element correctly use
$('#start') // or $('select#start')
It's better not to mess with HTML strings. You could use Option constructor to create new option element.
var locations = [{"name":"Hotel1","address":"Bondi Beach","lat":"6.369869","lng":"80.042055","thumb_path":"img\/hotel1.jpg"},{"name":"Hotel2","address":"Coogee Beach","lat":"6.025822","lng":"80.305727","thumb_path":"img\/hotel2.jpg"},{"name":"Hotel3","address":"Cronulla Beach","lat":"6.691994","lng":"79.928771","thumb_path":"img\/hotel3.jpg"},{"name":"Hotel4","address":"Manly Beach","lat":"6.887130","lng":"80.093420","thumb_path":"img\/hotel4.jpg"},{"name":"Hotel5","address":"Maroubra Beach","lat":"7.362609","lng":"79.825005","thumb_path":"img\/hotel5.jpg"}]
var select = $('#start')[0];
locations.forEach(function(location) {
select.add(new Option(location.name, location.name));
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<select id="start"></select>
Upvotes: 2