Question

我已经创建了两个警报，在成功的AddStaff上会出现成功警报，如果一个人员无法添加，则应出现警告警报。这就是我所做的：

    <div class="alert alert-success" style="display: none" id="AddLateStudentAlert"><div class="alert alert-warnning" style="display: none" id="WarningAlert">

这是我添加员工的ajax

function AddStaff(Staff) {
        ajReq.abort();
        ajReq = $.ajax({
            type: "POST",
            url: "Services/Staff.asmx/Staff",
            data: "{'Staff':'" + Staff+ "'}",
            contentType: "application/json; charset=utf-8",
            dataType: "json",

            success: function (msg) {

                $('#SucessAlert').show();
                $("#AddLateStudentAlert").fadeOut(5000);

            },else:(!sucess) {
                $('#WarningAlert').show();
    }

警报剂量不起作用，添加人员时，它始终显示警告警报。谁能告诉我为什么？

Answer 1

使用error，因为else不是$.ajax的有效功能。请参阅jQuery的 AJAX docs 。

示例：

success: function (msg) {

            $('#SucessAlert').show();
            $("#AddLateStudentAlert").fadeOut(5000);

        },
error: function(msg) {
            $('#WarningAlert').show();
}

Answer 2

function AddStaff(Staff) {
    ajReq.abort();
    ajReq = $.ajax({
        type: "POST",
        url: "Services/Staff.asmx/Staff",
        data: "{'Staff':'" + Staff+ "'}",
        contentType: "application/json; charset=utf-8",
        dataType: "json",

        success: function (msg) {

            $('#SucessAlert').show();
            $("#AddLateStudentAlert").fadeOut(5000);

        },
        error:function() {
            $('#WarningAlert').show();
        }
    }
}

Answer 3

jQuery.ajax.中没有else回调使用错误回调

       success: function (msg) {
            $('#SucessAlert').show();
            $("#AddLateStudentAlert").fadeOut(5000);
        },
       error:(jqXHR,textStatus,errorThrown ) {
              console.log(errorThrown)
            $('#WarningAlert').show();
       }

如果添加了记录，ajax会显示警报

3 个答案: