我期待顯示來自「conta」,「zip,city,street_a,street_b」,「count,number」的數據並從「contact」鍵創建一個下拉菜單,但這不會在哪裏,有人可以解釋這將如何工作,這是我得到了我所擁有的遠。從json數據創建下拉菜單
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>jQuery</title>
<!-- script type='text/javascript' href='https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js'></script -->
<script src="http://code.jquery.com/jquery-latest.js"></script>
<script type="text/javascript">
data = [
{"conta":"13356","name":"MISS Two"},
{"zip":"01111","city":"NASHUA, MA","street_a":"10 MAIN ROAD SQUARE","street_b":"Nothing"},
{"count":"544","number":"250"},
[
{"contact":"Mark Bre"},
{"contact":"Mary Lou"},
{"contact":"John Ton"},
{"contact":"Carls Des"},
{"contact":"Carlos Drt"}
]
]
$('body').append('<select id="dynamicSelect"></select>');
var options="";
$.each(data,function(i1,val1){
if($.isArray(val1)){
$.each(data,function(i2,val2){
options += "<option value="+ val2.contact +">" + val2.contact + "</option>"
});
}
}
$('#dynamicSelect').append(options);
</script>
</head>
<body>
The Select:<select id="dynamicSelect">
<option value="option value" selected="selected">option html</option>
</select>
</body>
</html>
明白並抓住數據的其他元素; 「conta」,「zip,city,street_a,street_b」,「count,number」它應該循環到數組中的哪個位置? – Andre 2013-03-19 00:57:28