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我正在尝试从 Placefinder API 检索信息。我的代码从文本框中获取信息，然后单击按钮将其发送到雅虎。 

            function codeAddress(){
            var address = document.getElementById("address").value;
            var RequestUrl = "http://where.yahooapis.com/geocode?q="+address+"&flags=J&callback=ws_results&output=json";
            JSONObject = new XMLHttpRequest();
            JSONObject.open( "GET", RequestUrl, false );
            JSONObject.send( null );
            return JSONObject;
            document.getElementByID("jlatitude").innerHTML=JSONObject.latitude;
            alert(document.write("jlatitude"));
        }

Firebug 告诉我数据已返回，但我无法在故障排除弹出窗口中显示我想要的内容。当我请求马里兰州的地理编码时，placefinder 会返回此信息。

{"ResultSet":{"version":"1.0","Error":0,"ErrorMessage":"No error","Locale":"us_US","Quality":10,"Found":2,"Results":[{"quality":49,"latitude":"39.466626","longitude":"-93.709069","offsetlat":"39.466626","offsetlon":"-93.709069","radius":700,"name":"","line1":"","line2":"Roads, MO","line3":"","line4":"United States","house":"","street":"","xstreet":"","unittype":"","unit":"","postal":"","neighborhood":"Roads","city":"Norborne","county":"Carroll County","state":"Missouri","country":"United States","countrycode":"US","statecode":"MO","countycode":"","uzip":"64668","hash":"","woeid":2482523,"woetype":7},{"quality":49,"latitude":"39.080130","longitude":"-82.537394","offsetlat":"39.080130","offsetlon":"-82.537394","radius":700,"name":"","line1":"","line2":"Roads, OH","line3":"","line4":"United States","house":"","street":"","xstreet":"","unittype":"","unit":"","postal":"","neighborhood":"Roads","city":"Wellston","county":"Jackson County","state":"Ohio","country":"United States","countrycode":"US","statecode":"OH","countycode":"","uzip":"45692","hash":"","woeid":2482522,"woetype":7}]}}

我的警报框没有出现，但数据已存储。

谢谢

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2 回答 2

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我认为您的 JavaScript 代码示例中有一个简单的问题 - 从 YQL 接收数据后，您有return JSONObject;这意味着以下行将永远不会被执行。

要进行故障排除和确认，您可以在返回时写入控制台日志，例如：

function codeAddress(){
    var address = document.getElementById("address").value;
    var RequestUrl = "http://where.yahooapis.com/geocode?q="+address+"&flags=J&callback=ws_results&output=json";
    JSONObject = new XMLHttpRequest();
    JSONObject.open( "GET", RequestUrl, false );
    JSONObject.send( null );
    console.log( JSONObject.responseText );
}
于 2012-08-14T22:44:43.480 回答
0 
$.getJSON(RequestUrl,{},function(result){
  ...
});

这会正确地从 Yahoo! 检索 JSON 对象！定位器

于 2012-08-17T16:38:25.310 回答