Question

我已经在网上阅读过有关此类错误的信息，并确定它并非来自多个jQuery核心源。我使用的是jQuery-1.10.2.min.js。

我正在处理的页面允许用户从列表中选择一个事件，之后应该在页面上显示属于该事件的活动列表。

jQuery / Ajax代码读取：

$("#event").change(function ()
{
    //get the selected value
    var eventId = $(this).val();
    alert( "sel val is " +  eventId);//this works

    $.ajax({
        dataType: 'json',
        type: 'GET',
        url: 'stcg-json-responses.php?fct=getJSONAllActivitiesAtEvent',
        data: "eventId="+eventId,
        cache: false,
        success: function(data)
        {
                alert("Data sent! " + eventId);//this works

                $.isArray(data.response)
                {
                    alert("this is an array");//this works
                    $.each(data.response, function (index, value)
                    {
                        // Loop through your results in jQuery and modify your HTML elements
                        htmlElement.append("<option id='key'>" + value + "</option>");
                    });
                }
        },
        error: function(xhr, status, error)
        {
            alert("Data not sent!");
            //console.log("Data not sent!");
        }
    });

使用各种浏览器开发工具查看结果，响应总是很好 - 它是以JSON对象的匿名数组的形式，每个对象以逗号分隔。但上面的脚本崩溃了： $ .each（data.response，function（index，value）

带有Uncaught TypeError。

为了让您更好地了解，响应数据的格式为（例如）：

[
{"activity_id":44,"activity_name":"Unknown activity","activity_desc":"Unknown activity","activity_short_code":"U","event_id":4,"capacity":0,"day_of_week":6,"date":"2012-03-24","project_leader":"","open":0},
{"activity_id":87,"activity_name":"Cleanup at Zoo Bellevue","activity_desc":"Zoo et parc d'accueil Bellevue","activity_short_code":"","event_id":4,"capacity":5,"day_of_week":6,"date":"2012-03-24","project_leader":"","open":0},
{"activity_id":89,"activity_name":"Ali Baba cleanup","activity_desc":"Ali Baba cleanup","activity_short_code":"","event_id":4,"capacity":10,"day_of_week":6,"date":"2012-03-24","project_leader":"","open":0}
]

任何人都知道这里发生了什么？非常感谢， swissphp

P.S。尝试使用普通的Javascript for循环，但它在array.length上崩溃。

=============================================== ===================================== 解决方案 - 基于此处收到的帮助，这是我的新代码，有效：

$("#event").change(function ()
{
    //get the selected value
    var eventId = $(this).val();

    var selectElement = $("#activity");

    $.ajax({
        dataType: 'json',
        type: 'GET',
        url: 'stcg-json-responses.php?fct=getJSONAllActivitiesAtEvent',
        data: "eventId="+eventId,
        cache: false,
        success: function(data)
        {
            if ($.isArray(data))
            {
                $.each(data, function (index, value)
                {
                    // Loop through your results in jQuery and modify your HTML elements
                    selectElement.append($('<option></option>').val(index).html(value['activity_name']));
                });
            }
        },
        error: function(xhr, status, error)
        { //etc etc etc

非常感谢所有贡献者！

Answer 1

您试图在$.each()上致电undefined，这会导致该错误。来自jQuery source：

each: function( obj, callback, args ) {
    var value,
        i = 0,
        length = obj.length,
        isArray = isArraylike( obj );

或者，换句话说，data.response未在服务器返回的JSON中定义。我怀疑问题在于你错过了if声明。

$.isArray(data.response)
{
    alert("this is an array");//this works
    $.each(data.response, function (index, value)
    {
        // Loop through your results in jQuery and modify your HTML elements
        htmlElement.append("<option id='key'>" + value + "</option>");
    });
}

这是一个毫无意义的声明（$.isArray(data.response)），后跟无条件的代码块。你可能意味着：

if($.isArray(data.response))
{
    alert("this is an array");//this works
    $.each(data.response, function (index, value)
    {
        // Loop through your results in jQuery and modify your HTML elements
        htmlElement.append("<option id='key'>" + value + "</option>");
    });
}

TypeError：无法读取undefined的属性'length' - 但是甚至不使用'length'！

1 个答案: