我怎样才能以编程方式获取纬度，位置的经度或使用api

Question

是获取位置的纬度或经度的任何方式。如果是，那么如何。

是使用谷歌地图API来做到这一点的任何方式。

Answer 1

“在浏览器中”一个简单的方法就是在点到中心的地图，你想粘贴以下的javascript到地址栏：

javascript:void(prompt('',gApplication.getMap().getCenter())); 

Answer 2

下面是如何做一个请求的示例使用JavaScript打开街道地图或Google地图。

// GOOGLE MAPS API v3 
// API: https://developers.google.com/maps/documentation/geocoding/#GeocodingRequests 
// Example JSON request: http://maps.googleapis.com/maps/api/geocode/json?address=1111%20W.%2035rd%20street,%20Chicago,%20IL&sensor=true 
function GoogleURI(address, type){ 
    var uri = "http://maps.googleapis.com/maps/api/geocode/"; 
    //address = FormatAddress(address); 
    if(type == "xml"){ 
    uri = uri + type + "?" + "address=" + address + "%26sensor=false"; 
    } else { // default to json 
    uri = uri + "json" + "?" + "address=" + address + "%26sensor=false"; 
    } 
    return uri; 
} 

// OPEN STREET MAP API 0.6 
// API: http://wiki.openstreetmap.org/wiki/Nominatim#Example 
// Example XML Request: http://nominatim.openstreetmap.org/search?q=%201111%20W.%2035th%20Street%2C%20Chicago%2C%20IL%2060609&format=xml&addressdetails=1 
// NOTE &'s and spaces dont pass easily to php and then to nominatim.openstreetmap.org 
// WARNING: OPEN STREET MAPS SOMETIMES DOESNT NEED THE STATE AND ZIP CODE and in fact will error out... ;) 
function OpenURI(address, type){ 
    var uri = "http://nominatim.openstreetmap.org/search?q="; 
    var format = "%26format=" + type; 
    var details = "%26addressdetails=1"; 
    //address = FormatAddress(address); 
    // NOTE: &'s dont pass good to php file_get_contents($uri) dont use "&polygon=1&addressdetails=1"; 
    if(type == "xml"){ 
    uri = uri + address + "%26format=" + type + "%26addressdetails=1"; 
    } else { // default to json 
    uri = uri + address + "%26format=" + "json" + "%26addressdetails=1"; 
    } 
    return uri; 
} 

你就可以发送从上述方法的返回URI，做在PHP中的GET请求。