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Javascript $.getJSON返回未识别的_Javascript_Json_Jquery_Google Maps Api 3 - Fatal编程技术网

Javascript $.getJSON返回未识别的

Javascript $.getJSON返回未识别的,javascript,json,jquery,google-maps-api-3,Javascript,Json,Jquery,Google Maps Api 3,我试图从GoogleMaps提供的JSON访问持续时间数据。用户输入2个邮政编码,I应显示所用时间 function getData() { var origin = $("#origin_pc").val() + ",australia"; var destination = $("#dest_pc").val() + ",australia"; // var origin="3135,australia"; // var destination="3155,a

我试图从GoogleMaps提供的JSON访问持续时间数据。用户输入2个邮政编码,I应显示所用时间

function getData() {
    var origin = $("#origin_pc").val() + ",australia";
    var destination = $("#dest_pc").val() + ",australia";
    // var origin="3135,australia";
    // var destination="3155,australia";
    url = "http://maps.googleapis.com/maps/api/directions/json?origin=" + origin + "&destination=" + destination + "&sensor=false";
    return url;
}

function calculateDistance() {
    // console.log(getData());
    var my_json;
    $.getJSON(getData(), function (json) {
        my_json = json;
    });
    console.log(my_json.routes[0].legs[0].duration.value);
}
这是我的HTML

<!DOCTYPE html>
<html>
    <head>
        <title></title>
    </head>

    <body>
        <form>
            <input type="text" id="origin_pc" />
            <input type="text" id="dest_pc" />
            <input type="submit" onclick="calculateDistance();">
        </form>
        <script type="text/javascript" src="jquery.js"></script>
        <script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false"></script>
        <script type="text/javascript" src="distance.js"></script>
    </body>
</html>
我对生成的URL进行了处理,打开后,它给出了正确的JSON对象。

return的786

您需要在getJSON回调函数中使用my_json

$.getJSON(getData(), function(json) {
   my_json = json;
   $('body').text(my_json.routes[0].legs[0].duration.value);
});