jquery each()项。[value]

时间:2014-09-04 17:58:55

标签: jquery uitableview key-value

<script type="text/javascript">
$(document).ready(function () {
    $("#CustomerIDbtn").click(function () {
        DoaTable(myTB, "../Customer/Index", "客户管理", "CustomerID", "Name", "#CustomerIDV", "#CustomerID")
    });
    $("#btnWBTypeID").click(function () {
        DoaTable(myTB, "../WBType/Index", "提单类型", "WBTypeID", "Name", "#WBTypeIDV", "#WBTypeID");
    });
    $("#btn").click(function () {
        DoaTable(myTB, "../Customer/GetCustomer");
    });
    function DoaTable(TableID, url,TLName,TID,TName,LV,LID) {
        $.ajax({
            url: url,
            type: "POST",
            dataType: "json",
            error: function () {
                alert("Error");
            },
            success: function (data) {
                var tbBody = "<tr><th>" + TLName + "</th></tr>";
                var a =  TID;
               $.each(data, function (i, item) {
                   tbBody += "<tr><td><a href=\"javascript:void(0)\"  class=\"aaf\" data-id=" + item.'TID' + " data-name=" + item.Name + ">" + item.Name + "</a></tr>";
                });
                $(TableID).empty();
                $(TableID).append(tbBody);
                $('.aaf').click(function () {
                    var usersid = $(this).data("id");
                    var userName = $(this).data("name");
                    $(LV).val(usersid);
                    $(LID).text(userName);
                })
            }
        });
    }
});

为什么TName不是equrt item.CustomerID它的问题是什么? 感谢你 如何才能错误地解决?

为什么item.TName不是equrt item.CustomerID?这是什么问题? 感谢你 如何才能错误地解决?

0 个答案:

没有答案