【问题标题】:how i can get latitude , longitude of a location programmatically or using a api我如何以编程方式或使用 api 获取位置的纬度、经度
【发布时间】:2010-09-01 09:22:41
【问题描述】:

是获取位置纬度或经度的任何方式。如果是,那么如何。

有任何方法可以使用 google map api 做到这一点。

【问题讨论】:

    标签: google-maps


    【解决方案1】:

    “在浏览器中”的一种简单方法是将地图居中放置在您想要的位置,然后将以下 javascript 粘贴到地址栏中:

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

    【讨论】:

      【解决方案2】:

      这是一个如何使用 JavaScript 为 Open Street Maps 或 Google Maps 发出请求的示例。

      // 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; 
      }
      

      然后您可以从上述方法发送返回 uri 并在 PHP 中执行 GET 请求。

      【讨论】:

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