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我有一个 .asmx 网络服务,它获取 cellid、MCC、MNC 和 LAC,我想使用 opencellid 从这些信息中获取纬度和经度,我没有在互联网上找到资源,所以我不知道在哪里开始。如果你能帮助我找出如何做到这一点,或者给我一个替代解决方案来从这些信息中找到纬度和经度,那就太好了。谢谢

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2 回答 2

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我面临同样的情况。读这个..答案就在这里。您可以将此代码合并到您的项目中..

http://android-coding.blogspot.com/2011/06/convert-celllocation-to-real-location.html

于 2012-10-21T03:26:18.970 回答
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如何使用GoogleMaps Api.

我在这里写了一篇关于它的博客文章:博客文章

将此类添加到您的项目中:

public class GoogleMapsApi
{
    static byte[] PostData(int MCC, int MNC, int LAC, int CID)
    {

        /* The shortCID parameter follows heuristic experiences:

         * Sometimes UMTS CIDs are build up from the original GSM CID (lower 4 hex digits)

         * and the RNC-ID left shifted into the upper 4 digits.

         */

        byte[] pd = new byte[] {

    0x00, 0x0e,

    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,

    0x00, 0x00,

    0x00, 0x00,

    0x00, 0x00,

    0x1b,

    0x00, 0x00, 0x00, 0x00, // Offset 0x11

    0x00, 0x00, 0x00, 0x00, // Offset 0x15

    0x00, 0x00, 0x00, 0x00, // Offset 0x19

    0x00, 0x00,

    0x00, 0x00, 0x00, 0x00, // Offset 0x1f

    0x00, 0x00, 0x00, 0x00, // Offset 0x23

    0x00, 0x00, 0x00, 0x00, // Offset 0x27

    0x00, 0x00, 0x00, 0x00, // Offset 0x2b

    0xff, 0xff, 0xff, 0xff,

    0x00, 0x00, 0x00, 0x00

};

        bool isUMTSCell = ((Int64)CID > 65535);

    /*    if (shortCID)

            CID &= 0xFFFF;      /* Attempt to resolve the cell using the

                        GSM CID part */

        if ((Int64)CID > 65536) /* GSM: 4 hex digits, UTMS: 6 hex

                        digits */

            pd[0x1c] = 5;

        else

            pd[0x1c] = 3;

        pd[0x11] = (byte)((MNC >> 24) & 0xFF);

        pd[0x12] = (byte)((MNC >> 16) & 0xFF);

        pd[0x13] = (byte)((MNC >> 8) & 0xFF);

        pd[0x14] = (byte)((MNC >> 0) & 0xFF);

        pd[0x15] = (byte)((MCC >> 24) & 0xFF);

        pd[0x16] = (byte)((MCC >> 16) & 0xFF);

        pd[0x17] = (byte)((MCC >> 8) & 0xFF);

        pd[0x18] = (byte)((MCC >> 0) & 0xFF);

        pd[0x27] = (byte)((MNC >> 24) & 0xFF);

        pd[0x28] = (byte)((MNC >> 16) & 0xFF);

        pd[0x29] = (byte)((MNC >> 8) & 0xFF);

        pd[0x2a] = (byte)((MNC >> 0) & 0xFF);

        pd[0x2b] = (byte)((MCC >> 24) & 0xFF);

        pd[0x2c] = (byte)((MCC >> 16) & 0xFF);

        pd[0x2d] = (byte)((MCC >> 8) & 0xFF);

        pd[0x2e] = (byte)((MCC >> 0) & 0xFF);

        pd[0x1f] = (byte)((CID >> 24) & 0xFF);

        pd[0x20] = (byte)((CID >> 16) & 0xFF);

        pd[0x21] = (byte)((CID >> 8) & 0xFF);

        pd[0x22] = (byte)((CID >> 0) & 0xFF);

        pd[0x23] = (byte)((LAC >> 24) & 0xFF);

        pd[0x24] = (byte)((LAC >> 16) & 0xFF);

        pd[0x25] = (byte)((LAC >> 8) & 0xFF);

        pd[0x26] = (byte)((LAC >> 0) & 0xFF);

        return pd;

    }

    static public string[] GetLatLng(int MCC, int MNC, int LAC, int CID)
    {

        string[] result = new string[2];

        try
        {

            String url = "http://www.google.com/glm/mmap";

            HttpWebRequest req = (HttpWebRequest)WebRequest.Create(

                new Uri(url));

            req.Method = "POST";

            byte[] pd = PostData(MCC, MNC, LAC, CID);

            req.ContentLength = pd.Length;

            req.ContentType = "application/binary";

            Stream outputStream = req.GetRequestStream();

            outputStream.Write(pd, 0, pd.Length);

            outputStream.Close();

            HttpWebResponse res = (HttpWebResponse)req.GetResponse();

            byte[] ps = new byte[res.ContentLength];

            int totalBytesRead = 0;

            while (totalBytesRead < ps.Length)
            {

                totalBytesRead += res.GetResponseStream().Read(

                    ps, totalBytesRead, ps.Length - totalBytesRead);

            }

            if (res.StatusCode == HttpStatusCode.OK)
            {

                short opcode1 = (short)(ps[0] << 8 | ps[1]);

                byte opcode2 = ps[2];

                int ret_code = (int)((ps[3] << 24) | (ps[4] << 16) |

                               (ps[5] << 8) | (ps[6]));

                if (ret_code == 0)
                {

                    double lat = ((double)((ps[7] << 24) | (ps[8] << 16)

                                 | (ps[9] << 8) | (ps[10]))) / 1000000;

                    double lon = ((double)((ps[11] << 24) | (ps[12] <<

                                 16) | (ps[13] << 8) | (ps[14]))) /

                                 1000000;

                    result[0] = lat.ToString();
                    result[1] = lon.ToString();
                    return result;

                }

                else
                {

                    return result;
                }

            }

            else

                return result;

        }

        catch (Exception ex)
        {
            result[0] = ex.Message;
            return result;

        }

    }
}

调用课程以获取您想要的纬度和对数:

string[] s = GoogleMapsApi.GetLatLng(MCC, MNC, lac, cellid);
double lat = 0;
Boolean b1 = double.TryParse(s[0], out lat);
double lon = 0;
Boolean b2 = double.TryParse(s[1], out lon);
于 2013-04-18T09:22:10.513 回答