如何使用Jquery ajax读取yahoo weather JSON数据

|| http://weather.yahooapis.com/forecastjson?w=2295424
{
\"units\":
{\"temperature\":\"F\",\"speed\":\"mph\",\"distance\":\"mi\",\"pressure\":\"in\"},
\"location\":{\"location_id\":\"INXX0075\",\"city\":\"Madras\",\"state_abbreviation\":\"*\",\"country_abbreviation\":\"IN\",\"elevation\":49,\"latitude\":13,\"longitude\":80.18000000000001},
\"wind\":{\"speed\":12.00000000000000,\"direction\":\"E\"},
\"atmosphere\":{\"humidity\":\"23\",\"visibility\":\"4.35\",\"pressure\":\"29.77\",\"rising\":\"steady\"},
\"url\":\"http:\\/\\/weather.yahoo.com\\/forecast\\/INXX0075.html\",\"logo\":\"http:\\/\\/l.yimg.com\\/a\\/i\\/us\\/nt\\/ma\\/ma_nws-we_1.gif\",\"astronomy\":{\"sunrise\":\"06:20\",\"sunset\":\"18:19\"},\"condition\":{\"text\":\"Sunny\",\"code\":\"32\",\"image\":\"http:\\/\\/l.yimg.com\\/a\\/i\\/us\\/we\\/52\\/32.gif\",\"temperature\":93.00000000000000},

\"forecast\":[{
\"day\":\"Today\",\"condition\":\"Mostly Clear\",\"high_temperature\":\"91\",\"low_temperature\":\"69\"},{\"day\":\"Tomorrow\",\"condition\":\"Partly Cloudy\",\"high_temperature\":\"90\",\"low_temperature\":\"70\"}
]}
我要显示“预测”
已邀请:
$.ajax({
   url: \"http://weather.yahooapis.com/forecastjson?w=2295424\",
   dataType: \"json\",
   success: function(data) {
      console.log( data.forecast[0].day );
      }
 });
在data.forecast [0] .day中,用所需的任何属性替换\“ day \”。
这应该是可行的:
console.debug(\"initYahooWeather starting\");
    var urlYahooWeather = \"http://weather.yahooapis.com/forecastjson?w=2295424\";

    var urlFlickr = \"http://weather.yahooapis.com/forecastjson?jsoncallback=?&w=2295424\";
    jQuery.ajax({
        async: false,
        type: \"POST\",
        contentType: \"application/json\",
        dataType: \"json\",
        url: urlFlickr,
        success: function(data){
            console.dir(data);
            console.dir(data.forecast);
            // here you have to navigate the array 

        },
        error: function(msg){
            console.debug(\"error contacting JSON server side component...\");
            console.debug(msg);
        }
    });

    console.debug(\"initYahooWeather stop\");
我尝试过自己,但是使用jQuery 1.5时收到一个解析错误:我不知道为什么,代码应该可以。 马济

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