紧接上篇，jQuery调用jsonp，并且在页面上展示

var Menu = function () {

return {
getMenuData: function (json) {
console.log(json);
var data = json.data;
var html = "";
for (var i = 0 ; i < data.length ; i ++) {
var url = data[i].u;
var name = data[i].n;
var sub = data[i].i;

html += "";
html += "<li class='dropdown-submenu'>";
html += "<a href='" + url + "'>" + name;
html += "<span class='c-arrow c-toggler'></span>";
html += "</a>";
html += "<ul class='dropdown-menu c-pull-right'>";

for (var j = 0 ; j < sub.length ; j ++) {
var url = sub[j].u;
var name = sub[j].n;
var node = sub[j].i;

html += "<li class='dropdown-submenu'>";
html += "<a href='" + url + "'>" + name;
html += "<span class='c-arrow c-toggler'></span>";
html += "</a>";

html += "<ul class='dropdown-menu c-pull-right'>";
for (var k = 0 ; k < node.length ; k ++) {
//                        debugger
var name = node[k];
var last = name.split("|");

html += "<li>";
html += "<a href='" + last[0] + "'>" + last[1] + "</a>";
html += "</li>";
}
html += "</ul>";
html += "</li>";
}

html += "</ul>";
html += "</li>";

}
$("#itemCatMenu").html(html);
},

getJSONP: function (serverUrl, callbackFun) {
$.ajax({
type: "get",
url: serverUrl,
dataType: "jsonp",
jsonp: "callback",
jsonpCallback: callbackFun,
success: function(json){
//                    console.log(json);
},
error: function(e){
if (e.status != "200") {
console.log(e);
}
}
});
}
};

}();

$(document).ready(function()
{
var serverUrl = "http://localhost:8088/rest/menu/list";
Menu.getJSONP(serverUrl, "Menu.getMenuData");
});