高分求救，服务端返回json问题，总是提示“invalid label”，大家帮帮忙！

Question

本帖最后由 zohong 于 2010-08-20 15:17:04 编辑

Answer 1

在Firebug中可以看到响应数据，但是提示错误“invalid label”
响应数据为：‘{"weatherinfo":{"city":"北京","city_en":"beijing","date_y":"2010年8月20日","date":"庚寅年七月十一","week":"星期五","fchh":"11","cityid":"101010100","temp1":"31℃~22℃","temp2":"26℃~20℃","temp3":"28℃~22℃","temp4":"30℃~21℃","temp5":"29℃~20℃","temp6":"25℃~18℃","tempF1":"87.8℉~71.6℉","tempF2":"78.8℉~68℉","tempF3":"82.4℉~71.6℉","tempF4":"86℉~69.8℉","tempF5":"84.2℉~68℉","tempF6":"77℉~64.4℉","weather1":"多云转阵雨","weather2":"阵雨转小到中雨","weather3":"阴转多云","weather4":"多云","weather5":"多云","weather6":"多云转中雨","img1":"1","img2":"3","img3":"3","img4":"21","img5":"2","img6":"1","img7":"1","img8":"99","img9":"1","img10":"99","img11":"1","img12":"8","img_single":"1","img_title1":"多云","img_title2":"阵雨","img_title3":"阵雨","img_title4":"小到中雨","img_title5":"阴","img_title6":"多云","img_title7":"多云","img_title8":"多云","img_title9":"多云","img_title10":"多云","img_title11":"多云","img_title12":"中雨","img_title_single":"多云","wind1":"微风","wind2":"微风","wind3":"微风","wind4":"微风","wind5":"微风","wind6":"微风","fx1":"微风","fx2":"微风","fl1":"小于3级","fl2":"小于3级","fl3":"小于3级","fl4":"小于3级","fl5":"小于3级","fl6":"小于3级","index":"炎热","index_d":"天气炎热，建议着短衫、短裙、短裤、薄型T恤衫、敞领短袖棉衫等清凉夏季服装。","index48":"热","index48_d":"天气较热，建议着短裙、短裤、短套装、T恤等夏季服装。年老体弱者宜着长袖衬衫和单裤。","index_uv":"中等","index48_uv":"弱","index_xc":"不宜","index_tr":"适宜","index_co":"较不舒适","st1":"31","st2":"21","st3":"22","st4":"19","st5":"30","st6":"22","index_cl":"适宜","index_ls":"适宜"}}’

Answer 2

格式有问题[{city:"values",weatherinfo:"values"}]

Answer 3

$.getJSON("/Service/ObtainHouseSource.html", { "x1": x1, "x2": x2, "y1": y1, "y2": y2, }, function(data)
    {
    $.each(data, function(i, item)
        {
alert(item.LocationLat);
        });
    });
js端

Answer 4

 
var josn=eval(data);
 for (var i = 0; i < json.weatherinfo.length; i++)
{
  alert(json.JsonData[i].name);
}这样你试试

Answer 5

  
var josn=eval(data);
 for (var i = 0; i < json.weatherinfo.length; i++)
{
  alert(json.weatherinfo[i].name);
}这样你试试

Answer 6

  
var josn=eval(data);
 for (var i = 0; i < json.weatherinfo.length; i++)
{
  alert(json.weatherinfo[i].city);
}最后个对的吧，这是JS的写法，

Answer 7

是格式问题json ：
{"weatherinfo":[{"city":"北京","city_en":"beijing","date_y":"2010年8月20日","date":"庚寅年七月十一","week":"星期五","fchh":"11",...},{......}]

Answer 8

在firebug中转换可以的
用eval('(json字符串)');//返回object
----------
>>> eval('({"weatherinfo":{"city":"北京","city_en":"be...st6":"22","index_cl":" 不宜","index_ls":"不宜"}})');
Object { weatherinfo=Object}
-------------
用eval('[json字符串]');//返回array
---------
>>> eval('[{"weatherinfo":{"city":"北京","city_en":"be...st6":"22","index_cl":" 不宜","index_ls":"不宜"}}]');
[Object { weatherinfo=Object}]

Answer 9

还没等我处理数据的时候，就提示了“invalid label”这个错误。就是那个代码，我要怎么该一下他就不会出这个错了。高手帮忙动个手术，服务端改不了，是中国天气网的数据。。

Answer 10

我知道答案。跨域不是这样的，你这个肯定不能实现。
你打开 细读 json 官方文档后，翻到最后，你会发现这么一段： JSON数据是一种能很方便通过JavaScript解析的结构化数据。如果获取的数据文件存放在远程服务器上（域名不同，也就是跨域获取数据），则需要使用jsonp类型。使用这种类型的话，会创建一个查询字符串参数 callback=? ，这个参数会加在请求的URL后面。服务器端应当在JSON数据前加上回调函数名，以便完成一个有效的JSONP请求。如果要指定回调函数的参数名来取代默认的callback，可以通过设置$.ajax()的jsonp参数。 其实jquery跨域的原理是通过外链 <script>  来实现的,然后在通过回调函数加上回调函数的参数来实现真正的跨域 Jquery 在每次跨域发送请求时都会有callback这个参数，其实这个参数的值就是回调函数名称，所以，服务器端在发送json数据时，应该把这个参数放到前面，这个参数的值往往是随机生成的，如：jsonp1294734708682，同时也可以通过 $.ajax 方法设置 callback 方法的名称。明白了原理后，服务器端应该这样发送数据：string message = "jsonp1294734708682({\"userid\":0,\"username\":\"null\"})";这样，json 数据 {\"userid\":0,\"username\":\"null\"} 就作为了 jsonp1294734708682 回调函数的一个参数所以，跨域的页面，你也要能控制啊。