谷歌地图地理编码 API V3 未在 javascript 函数中返回结果

Google maps geocode API V3 not returning result in javascript function

提问by Jason

I'm trying to use the Google geocoder API V3 to plot a location on a map based on an address specified by the user, code is below.

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。

When I make a request directly (e.g. to http://maps.googleapis.com/maps/api/geocode/json?address=peterborough&sensor=false) I get the expected response. However, when I make the same request using the code below, the midpointvariable is always undefined after the getLatLongfunction has exited.

当我直接提出请求（例如http://maps.googleapis.com/maps/api/geocode/json?address=peterborough&sensor=false）时，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。

What am I doing incorrectly?

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;
}

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

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

In light of responses, I have now updated the code to the following. I'm still not getting any result though, with breakpoints set in Firebug the result = data.results[0].geometry.location;never gets hit.

根据响应，我现在已将代码更新为以下内容。我仍然没有得到任何结果，在 Firebug 中设置了断点result = 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);
  ...
}

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

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

I have it! The final code, which works, looks like this:

我有！最终有效的代码如下所示：

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);
  ...
}

回答by Scott

If you are using V3 of the API cannot you use the this?

如果您使用的是 V3 的 API，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);
        }
    });
}

The above is what I use to find some lat long cordinates of an inputted address, May work better?

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？

EDIT:

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;
}

回答by DanSingerman

The problem is your $.getJSONfunction is asynchronous, yet you are returning the 'result' synchronously.

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

You need to do something like this (not tested!)

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 
    }
  });
}

In response to your edit: Oh dear, it looks like the V3 geocoder does not support JSONP. This means you can not do a cross-domain request to get data from it in your browser. See http://blog.futtta.be/2010/04/09/no-more-jsonp-for-google-geocoding-webservice/

回应您的编辑：哦，亲爱的，V3 地理编码器似乎不支持 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/

However Brady's solution does work. I guess that is the way Google want us to geocode now.

但是 Brady 的解决方案确实有效。我想这就是 Google 现在希望我们进行地理编码的方式。