我如何更快地发送AJAX请求

时间:2016-03-16 19:14:57

标签: javascript jquery ajax

我想同时发送AJAX请求。我尝试使用for循环

for(i = 0;i <= 10; i ++){
    $.get("http://search.roblox.com/catalog/json?CatalogContext=1&SortType=0&SortAggregation=3&SortCurrency=0&LegendExpanded=true&Category=0", function(r){
        var date = new Date();
        console.log(date.getTime());
    });
}

但是,它们不会同时发送,并且记录的时间之间存在明显差异。我想基本上在彼此的同时发送它们(如果可能的话)。 我想知道是否有任何方法可以做这样的事情?

1 个答案:

答案 0 :(得分:-4)

 <!DOCTYPE html>
@*!!!!!!!!!! added to routes routes.IgnoreRoute("{resource}.axd/{*pathInfo}");!!!!!!!*@
<html>
<head>
    <meta name="viewport" content="width=device-width" />
    <title>Index</title>
    <script src="~/Scripts/jquery-1.8.2.min.js"></script>
    <style type="text/css">
        body {
            width: 300px;
        }
    </style>
</head>
<body>


    <div class="aSpinner" style="display: none;" onclick="">
        <img src="~/Content/ajax_busy2.gif" title="Please wait.... Contacting server for action..." />
        ...Please wait.... Contacting server for action...
    </div>
    <div id="divMore"></div>

    <div>
        <script type="text/javascript">

            $(".aSpinner").show();

             $.when(
                    $.get("/Main/Index01", function (html01) {
                        $("#divMore").append(html01);
                    }),
                    $.get("/Main/Index02", function (html02) {
                        $("#divMore").append(html02);
                    }),
                    $.get("/Main/Index03")
                        ).then(function () {
                            // All is ready now, so...
                            $("#divMore").append("all three call done aynchronously");
                        });
        </script>


    </div>
</body>
</html>

ously“);

    public class MainController : Controller
    {
        //
        // GET: /Main/

        public string Index01()
        {
            return "index01";
        }

        public string Index02()
        {
            return "index02";
        }

        public string Index03()
        {
            return "index03";
        }

enter image description here                         });