c# - 如何发布谷歌距离矩阵参数？

Question

Google 距离矩阵最多可以接受 100 个参数。但是使用 GET Ruquest url lenth 将其限制为 < 15（我认为 2048 个字符），并且我得到了错误 414 - 请求的 URL 太大而无法处理。因此我得出结论，必须使用 POST 方法。但我不能。我收到了 REQUEST_DENIED 错误。那么我该如何使用这项服务呢？

    public static bool GetMatrix(string origins, string destinations)
    {
        string poststring = string.Format("origins={0}&destinations={1}&mode=bicycling&language=fr-FR&sensor=false", origins, destinations);
        byte[] postdata = Encoding.UTF8.GetBytes(poststring);

        HttpWebRequest webRequest = (HttpWebRequest)WebRequest.Create("http://maps.googleapis.com/maps/api/distancematrix/xml");
        webRequest.Method = "POST";
        webRequest.ContentType = "application/xml"; // or any other type dont work
        webRequest.ContentLength = postdata.Length;
        using (Stream writer = webRequest.GetRequestStream())
            writer.Write(postdata, 0, postdata.Length);

        using (HttpWebResponse webResponse = (HttpWebResponse)webRequest.GetResponse())
        {
            //Only for debug
            using (var stream = new StreamReader(webResponse.GetResponseStream()))
                System.Diagnostics.Debug.WriteLine(stream.ReadToEnd());

            return (webResponse.StatusCode == HttpStatusCode.OK);
        }
    }

Answer 1

JS方法

例如。距离矩阵示例

var origin1 = new google.maps.LatLng(55.930385, -3.118425);
var origin2 = "Greenwich, England";
var destinationA = "Stockholm, Sweden";
var destinationB = new google.maps.LatLng(50.087692, 14.421150);

var service = new google.maps.DistanceMatrixService();
service.getDistanceMatrix(
  {
    origins: [origin1, origin2],
    destinations: [destinationA, destinationB],
    travelMode: google.maps.TravelMode.DRIVING,
    avoidHighways: false,
    avoidTolls: false
  }, callback);

function callback(response, status) {
  // See Parsing the Results for
  // the basics of a callback function.
}

C# 方法在没有 google.com 的情况下计算距离

 public decimal calcDistance(decimal latA, decimal longA, decimal latB, decimal longB)
    {

        double theDistance = (Math.Sin(DegreesToRadians(latA)) *
                Math.Sin(DegreesToRadians(latB)) +
                Math.Cos(DegreesToRadians(latA)) *
                Math.Cos(DegreesToRadians(latB)) *
                Math.Cos(DegreesToRadians(longA - longB)));

        return Convert.ToDecimal((RadiansToDegrees(Math.Acos(theDistance)))) * 69.09M * 1.6093M;
    }