我有一个JSON代码:
"rows": [
{
"id": 1,
"name": "nazwa1"
},
{
"id": 2,
"name": "nazwa2"
},
{
"id": 3,
"name": "nazwa3"
},
...
{
"id": 99,
"name": "nazwa99"
}
]
在jQuery中,我在listName
脚本如下:
$('#displayData').dataTable({
"aaData": listName,
"aoColumns": [
{ "mDataProp": "id" },
{ "mDataProp": "name" }
]
});
但是错了,如何从listName中完成jQuery中的数据?
答案 0 :(得分:1)
fxnRelatedData = function () {
$(document).ready(function () {
$('#displayData').change(function () {
$.each(rows, function (index) {
/* give some class name to your table rows*/
$(".ClassName").html(rows[index].name);
});
});
});
};
fxnRelatedData();