getJson 并解析以创建列表（目前） jQuery/Yahoo Finance API

发布于 2024-12-20 08:48:38 字数 851 浏览 1 评论 0原文

我几乎可以正常工作了，但我无法解决解析问题。如果有人可以帮助我将非常感激！

我正在尝试查询 Yahoo Finance API 并使用 jQuery 解析结果。这是我执行此操作的代码：

  <script>
$(document).ready(function(){

var url = "http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.quotes%20WHERE%20symbol%3D'NPO'&format=json&diagnostics=true&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys&callback";

$.getJSON(url + "&format=json&jsoncallback=?", function(data) {
  var items = [];

  $.each(data, function(key, val) {
    items.push('<li id="' + key + '">' + val + '</li>');
  });

  $('<ul/>', {
    'class': 'my-new-list',
    html: items.join('')
  }).appendTo('body');
});


});
</script>

但我收到此错误：在此处输入图像描述

任何克服此错误的帮助将不胜感激。

谢谢！

原文

I almost have the thing working but I can't get past a parsing issue. If anyone can help I would be very thankful!

I am trying to query Yahoo Finance API and parse the results using jQuery. Here is my code to do so:

  <script>
$(document).ready(function(){

var url = "http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.quotes%20WHERE%20symbol%3D'NPO'&format=json&diagnostics=true&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys&callback";

$.getJSON(url + "&format=json&jsoncallback=?", function(data) {
  var items = [];

  $.each(data, function(key, val) {
    items.push('<li id="' + key + '">' + val + '</li>');
  });

  $('<ul/>', {
    'class': 'my-new-list',
    html: items.join('')
  }).appendTo('body');
});


});
</script>

But I am getting this error:
enter image description here

Any help overcoming this error would be much appreciated.

Thanks!

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

下雨或天晴 2024-12-27 08:48:38

服务未使用 jsoncallback 参数。
您甚至不必指定回调参数。它是通过 getJSON() 添加的
format 参数已在 url 中指定
您的 items 数组将对象存储为数据位于 data.query.results.quote 下

试试这个：

var url = "http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.quotes%20WHERE%20symbol%3D'NPO'&format=json&diagnostics=true&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys";

$.getJSON(url, function(data) {
  var items = [];
  $.each(data.query.results.quote, function(key, val) {
    items.push('<li id="' + key + '">' + val + '</li>');
  });
  $('<ul/>', { 'class': 'my-new-list', html: items.join('')}).appendTo('body');
});

工作代码是这里。

jsoncallback parameter is not used by the service.
You even don't have to specify callback parameter. It is added by the getJSON()
format parameter is already specified in url
Your items array is storing objects as the data are under data.query.results.quote

Try this:

var url = "http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.quotes%20WHERE%20symbol%3D'NPO'&format=json&diagnostics=true&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys";

$.getJSON(url, function(data) {
  var items = [];
  $.each(data.query.results.quote, function(key, val) {
    items.push('<li id="' + key + '">' + val + '</li>');
  });
  $('<ul/>', { 'class': 'my-new-list', html: items.join('')}).appendTo('body');
});

Working code is HERE.

回复收藏 0 原文

怀里藏娇 2024-12-27 08:48:38

YQL 使用 callback=? 参数，而不是 jsoncallback=? 尝试以下操作：

var url = "http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.quotes%20WHERE%20symbol%3D'NPO'&format=json&diagnostics=true&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys";

$.getJSON(url + "&format=json&callback=?", function(data) {

编辑：注意，url 也必须更改。

YQL uses a callback=? parameter, not jsoncallback=? try this:

var url = "http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.quotes%20WHERE%20symbol%3D'NPO'&format=json&diagnostics=true&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys";

$.getJSON(url + "&format=json&callback=?", function(data) {

Edit: Note, the url had to change too.

回复收藏 0 原文

不弃不离 2024-12-27 08:48:38

当您删除 &jsoncallback=? 时，对我来说效果很好。

$(document).ready(function() {

    var url = "http://query.yahooapis.com/v1/public/yql?" +                    
              "q=select%20*%20from%20yahoo.finance.quotes%20WHERE%20symbol%3D'NPO'" + 
              "&format=json&diagnostics=true" + 
              "&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys";

    $.getJSON(url, function(data) {

        console.log( data );

    });
});

实际上，主字符串中已经有了 format=json 。

JSFIDDLE 演示

Works fine for me when you remove the &jsoncallback=?.

$(document).ready(function() {

    var url = "http://query.yahooapis.com/v1/public/yql?" +                    
              "q=select%20*%20from%20yahoo.finance.quotes%20WHERE%20symbol%3D'NPO'" + 
              "&format=json&diagnostics=true" + 
              "&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys";

    $.getJSON(url, function(data) {

        console.log( data );

    });
});

You actually already have the format=json in the main string.

JSFIDDLE DEMO

回复收藏 0 原文

~没有更多了~