我已经决定了我的JSON，来自服务器，是有效的（手工制作Ajax调用），但我真的想使用jQuery。我也确定，后的网址，被发送到服务器，是正确的，使用萤火虫。但是，错误回调仍然被触发（parsererror）。我也试过数据类型：文本

是否还有其他的选择，我应该包括哪些内容？

  $（函数（）{
    $（＃提交）。绑定（点击，函数（）{
        $阿贾克斯（{
            类型：后，
            网址：HTTP：// MYSERVER / cgi-bin目录/经纪人，
            数据类型：JSON，
            数据：{启动：开始，结束：结束}，
            错误：函数（请求错误）{
                警报（错误）;
            }，
            成功：函数（要求）{
                警报（request.length）;
            }
        }）; //结束阿贾克斯
    }）; //结束绑定
}）; //结束事件监听器
 

解决方案

下面是一些建议，我会尝试：

1）已指定要的dataType的数据类型选项（区分大小写的，我相信）

2）尽量使用'的contentType'选项，因为这样：

 的contentType：应用/ JSON的;字符集= UTF-8
 

我不知道有多少，这将帮助，因为它是用在要求你的帖子的网址，而不是在响应中。 看到这篇文章的详细信息：http://encosia.com/2008/06/05/3-mistakes-to-avoid-when-using-jquery-with-aspnet-ajax （这对asp.net编写的，但可以适用）

3）三重检查你的帖子的网址的输出和运行通过JSON验证器的输出只是要绝对确保它是有效的，并且可以被解析成一个JSON对象。 http://www.jsonlint.com

希望一些这方面的帮助！

I have determined that my JSON, coming from the server, is valid (making the ajax call manually), but I would really like to use JQuery. I have also determined that the "post" url, being sent to the server, is correct, using firebug. However, the error callback is still being triggered (parsererror). I also tried datatype: text.

Are there other options that I should include?

$(function() {
    $("#submit").bind("click", function() {
        $.ajax({
            type: "post",
            url: "http://myServer/cgi-bin/broker" ,
            datatype: "json",
            data: {'start' : start,'end' : end},
            error: function(request,error){
                alert(error);
            },
            success: function(request) {
                alert(request.length);
            }
        }); // End ajax
    }); // End bind
}); // End eventlistener

Here are a few suggestions I would try:

1) the 'datatype' option you have specified should be 'dataType' (case-sensitive I believe)

2) try using the 'contentType' option as so:

contentType: "application/json; charset=utf-8"

I'm not sure how much that will help as it's used in the request to your post url, not in the response. See this article for more info: http://encosia.com/2008/06/05/3-mistakes-to-avoid-when-using-jquery-with-aspnet-ajax (It's written for asp.net, but may be applicable)

3) Triple check the output of your post url and run the output through a JSON validator just to be absolutely sure it's valid and can be parsed into a JSON object. http://www.jsonlint.com

Hope some of this helps!