5

我正在尝试使用 Google geocoder API V3 根据用户指定的地址在地图上绘制位置,代码如下。

当我直接发出请求时(例如向http://maps.googleapis.com/maps/api/geocode/json?address=peterborough&sensor=false),我得到了预期的响应。但是,当我使用下面的代码发出相同的请求时,在getLatLong函数退出后,中点变量始终未定义。

我做错了什么?

function loadFromSearch(address) 
{
  midpoint = getLatLong(address);
  mapCentre = midpoint;
  map.setMapTypeId(google.maps.MapTypeId.ROADMAP);
  ...
}


function getLatLong(address) 
{
  var result;
  var url = 'http://maps.googleapis.com/maps/api/geocode/json?address=' + encodeURIComponent(address) + '&sensor=false'
  $.getJSON(url,
  function (data){
     if (data.status == "OK")
     {
        result = data.results[0].geometry.location;
     }
  });
  return result;
}

==================================================== =================================

根据响应,我现在已将代码更新为以下内容。不过,我仍然没有得到任何结果,在 Firebug 中设置了断点,结果 = data.results[0].geometry.location; 永远不会被击中。

function loadFromSearch(address) 
{
  midpoint = getLatLong(address, loadWithMidpoint);    
}

function getLatLong(address, callback)
{
   var result;
   var url = 'http://maps.googleapis.com/maps/api/geocode/json?address=' + encodeURIComponent(address) + '&sensor=false'
   $.getJSON(url,{},
   function (data) {
     if (data.status == "OK")
     {
        result = data.results[0].geometry.location;
        callback(result);
     }
   });
}

function loadWithMidpoint(centre)
{
  mapCentre = centre;
  map.setMapTypeId(google.maps.MapTypeId.ROADMAP);
  ...
}

==================================================== ============================

我有!有效的最终代码如下所示:

function loadFromSearch(coordinates, address)
{
  midpoint = getLatLong(address, latLongCallback);
}

function getLatLong(address, callback)
{
  var geocoder = new google.maps.Geocoder();
  var result = "";
  geocoder.geocode({ 'address': address, 'region': 'uk' }, function (results, status) {
     if (status == google.maps.GeocoderStatus.OK)
     {
        result = results[0].geometry.location;
        latLongCallback(result);
     }
     else
     {
        result = "Unable to find address: " + status;
     }
  });
  return result;
}

function latLongCallback(result)
{
  mapCentre = result;
  map.setMapTypeId(google.maps.MapTypeId.ROADMAP);
  ...
}
4

2 回答 2

10

如果您使用 API 的 V3,您不能使用这个吗?

function findAddressViaGoogle() {
    var address = $("input[name='property_address']").val();
    var geocoder = new google.maps.Geocoder();
    geocoder.geocode( { 'address': address, 'region': 'uk' }, function(results, status) {
        if (status == google.maps.GeocoderStatus.OK) {
            newPointClicked(results[0].geometry.location)
        } else {
            alert("Unable to find address: " + status);
        }
    });
}

以上是我用来查找输入地址的一些经纬度坐标的方法,可能会更好吗?

编辑:

function loadFromSearch(address) 
{
midpoint = getLatLong(address);
mapCentre = midpoint;
map.setMapTypeId(google.maps.MapTypeId.ROADMAP);
...
}


function getLatLong(address) 
{
    var geocoder = new google.maps.Geocoder();
    var result = "";
    geocoder.geocode( { 'address': address, 'region': 'uk' }, function(results, status) {
        if (status == google.maps.GeocoderStatus.OK) {
            result = results[0].geometry.location;
        } else {
            result = "Unable to find address: " + status;
        }
    });
    return result;
}
于 2010-11-09T10:41:06.187 回答
4

问题是您的$.getJSON函数是异步的,但您正在result同步返回“”。

你需要做这样的事情(未经测试!)

function loadFromSearch(address) 
{
  midpoint = getLatLong(address, function(midpoint){
    // this is a callback
    mapCentre = midpoint;
    map.setMapTypeId(google.maps.MapTypeId.ROADMAP);
    ...           
    });
}

function getLatLong(address, callback) 
{
var result;
var url = 'http://maps.googleapis.com/maps/api/geocode/json?address=' + encodeURIComponent(address) + '&sensor=false'
$.getJSON(url,
  function (data) {
    if (data.status == "OK") {
        result = data.results[0].geometry.location;
        callback(result) // need a callback to get the asynchronous request to do something useful 
    }
  });
}

回应您的编辑:哦,天哪,看起来 V3 地理编码器不支持 JSONP。这意味着您无法在浏览器中执行跨域请求以从中获取数据。见http://blog.futtta.be/2010/04/09/no-more-jsonp-for-google-geocoding-webservice/

然而,布雷迪的解决方案确实有效。我想这就是谷歌希望我们现在进行地理编码的方式。

于 2010-11-09T10:44:23.147 回答