地址已知时如何获得坐标？

Question

地址已知时如何获得坐标？

B. *_*non 12 c# geocoding geolocation openstreetmap windows-store-apps

我从Adam Freeman的书"Metro Revealed:使用XAML和C#构建Windows 8应用程序"中获取/调整了以下代码,以便在知道坐标时获取地址:

public static async Task<string> GetAddressForCoordinates(double latitude, double longitude)
{
    HttpClient httpClient = new HttpClient {BaseAddress = new Uri("http://nominatim.openstreetmap.org")};
    HttpResponseMessage httpResult = await httpClient.GetAsync(
        String.Format("reverse?format=json&lat={0}&lon={1}", latitude, longitude));
JsonObject jsonObject = JsonObject.Parse(await httpResult.Content.ReadAsStringAsync());

return jsonObject.GetNamedObject("address").GetNamedString("road");

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}

如何获得相反的结果(如果已知地址,则为坐标)？

UPDATE

我正在为此增加一笔赏金; 我已经得到的(如上所示)是反向地理编码(获取坐标的地址); 我需要的是地理编码(获取地址的坐标).

基于我上面的反向地理编码代码,我猜它可能是这样的:

public static async Task<string> GetCoordinatesForAddress(string address)
{
    HttpClient httpClient = new HttpClient {BaseAddress = new Uri("http://nominatim.openstreetmap.org")};
    HttpResponseMessage httpResult = await httpClient.GetAsync(
        String.Format("format=json&address={0}", address));

    JsonObject jsonObject = JsonObject.Parse(await httpResult.Content.ReadAsStringAsync());

    return jsonObject.GetNamedObject("address").GetNamedString("lat"); // <-- what about "lon"?
}

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...但我不知道如何组合两个坐标(经度和纬度)值(假设这是正确的,或接近正确).任何人都可以验证这一点,清理它,或者提供一个更好的例子(使用nominatim或其他方式)？

更新2

要回答Peter Ritchie的问题/评论如下:

在原始(反向地理编码代码)中,我得到了:

return jsonObject.GetNamedObject("address").GetNamedString("road");

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它只是回归道路; 所以我认为像"157 Riverside Avenue"这样的东西.

但对于地理编码(需要两个值,经度和纬度),我有这个伪代码:

return jsonObject.GetNamedObject("address").GetNamedString("lat"); // <-- what about "lon"?

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所以我不知道是否需要将返回值从Task <string> 更改为Task <List并调用(verbose伪代码)[注意:我很难通过字符串列表转义任务的尖括号]:

var latitude jsonObject.GetNamedObject("address").GetNamedString("lat");
var longitude jsonObject.GetNamedObject("address").GetNamedString("lat");
List<string> listCoordinates = new List<string>();
listCoordinates.Add(latitude);
listCoordinates.Add(longitude);
return listCoordinates;

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......或者像这样:

string latitude jsonObject.GetNamedObject("address").GetNamedString("lat");
string longtude jsonObject.GetNamedObject("address").GetNamedString("long");
return string.Format("{0};{1}", latitude, longitude);

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...要么？？？

更新3

响应提供的Json代码进行地理编码:

基于原始的反向地理编码代码,调用不应该更像这样:

HttpClient httpClient = new HttpClient { BaseAddress = new Uri("http://nominatim.openstreetmap.org/") };
var httpResult = await httpClient.GetAsync(
    String.Format("search?format=json&addressdetails={0}", address);

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...但无论如何:虽然JsonArray是JArray类型,但JArray类型无法识别.虽然JsonValue是,但无法识别JValue类型.JsonConverter类型无法识别; 也许是Json.Net的一部分？

我最接近编译的proferred代码是:

var result = await httpResult.Content.ReadAsStringAsync();
var r = (JsonArray)JsonConverter.DeserializeObject(result);//<-- JsonConvert[er] not recognized; part of Json.NET?
var latString = ((JsonValue)r[0]["lat"]).ValueType as string;
var longString = ((JsonValue)r[0]["lon"]).ValueType as string;

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...但即便如此(关闭但没有Bob Seger),JsonConvert和JsonConverter都无法识别.

更新4

在通过http://wiki.openstreetmap.org/wiki/Nominatim#Search上的文档协调一致地进行了更多的讨论后,我认为我原来的(反向地理编码)方法可能更好:

public static async Task`<string`> GetAddressForCoordinates(double latitude, double longitude)
{
    HttpClient httpClient = new HttpClient {BaseAddress = new Uri("http://nominatim.openstreetmap.org/")};
    HttpResponseMessage httpResult = await httpClient.GetAsync(
        String.Format("reverse?format=json&lat={0}&lon={1}", latitude, longitude));

    JsonObject jsonObject = JsonObject.Parse(await httpResult.Content.ReadAsStringAsync());

    string house = jsonObject.GetNamedObject("addressparts").GetNamedString("house");
    string road = jsonObject.GetNamedObject("addressparts").GetNamedString("road");
    string city = jsonObject.GetNamedObject("addressparts").GetNamedString("city");
    string state = jsonObject.GetNamedObject("addressparts").GetNamedString("state");
    string postcode = jsonObject.GetNamedObject("addressparts").GetNamedString("postcode");
    string country = jsonObject.GetNamedObject("addressparts").GetNamedString("country");
    return string.Format("{0} {1}, {2}, {3} {4} ({5})", house, road, city, state, postcode, country);
}

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对于相应的坐标传递,这将返回,例如:" 157 Riverside Avenue,Champaign,IL 55555(USA) "

我对文档的奇怪之处在于地址部分中没有"状态"元素; 如果确实如此,而不仅仅是文档疏忽,我上面的代码在调用GetNamedString("state")时会失败.

我仍然不确定对于相反的(地理编码)方法应该使用正确的语法等,在传入地址后获取坐标.

更新5

好的,我下载了Json.NET并进行了编译.我还没有测试过,但我已经将Peter Ritchie标记为(50分)答案.

这是我正在使用的代码:

public static async Task<string> GetCoordinatesForAddress(string address)
{
    HttpClient httpClient = new HttpClient { BaseAddress = new Uri("http://nominatim.openstreetmap.org/") };
    HttpResponseMessage httpResult = await httpClient.GetAsync(
        String.Format("search?q={0}&format=json&addressdetails=1", Pluggify(address))); // In my Pluggify() method, I replace spaces with + and then lowercase it all

    var result = await httpResult.Content.ReadAsStringAsync();
    var r = (JArray)JsonConvert.DeserializeObject(result);
    var latString = ((JValue)r[0]["lat"]).Value as string;
    var longString = ((JValue)r[0]["lon"]).Value as string;
    return string.Format("{0};{1}", latString, longString);
}

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另外:在回到这个论坛的路上发生了一件有趣的事情:在通过NuGet安装Json.NET时,我还看到了"ServiceStack的.NET最快的JSON Serializer",它声称比Json.NET快3倍.它最近比Json.NET更新了.思考/反应？

更新6

我有这个代码来实现这个(app id和代码已被更改以保护半无辜(我)):

// If address has not been explicitly entered, try to suss it out:
                    address = textBoxAddress1.Text.Trim();
                    lat = textBoxLatitude1.Text.Trim();
                    lng = textBoxLongitude1.Text.Trim();
                    if (string.IsNullOrWhiteSpace(address))
                    {
                        address = await SOs_Classes.SOs_Utils.GetAddressForCoordinates(lat, lng);
                    }

. . .

        public async static Task<string> GetAddressForCoordinates(string latitude, string longitude)
        {
            string currentgeoLoc = string.Format("{0},{1}", latitude, longitude);
            string queryString = string.Empty;
            string nokiaAppID = "j;dsfj;fasdkdf";
            object nokiaAppCode = "-14-14-1-7-47-178-78-4";
            var hereNetUrl = string.Format(
                "http://demo.places.nlp.nokia.com/places/v1/discover/search?at={0}&q={1}&app_id={2}    
&app_code={3}&accept=application/json",
                    currentgeoLoc, queryString, nokiaAppID, nokiaAppCode);    
            // get data from HERE.net REST API
            var httpClient = new HttpClient();
            var hereNetResponse = await httpClient.GetStringAsync(hereNetUrl);    
            // deseralize JSON from Here.net 
            using (var tr = new StringReader(hereNetResponse))
            using (var jr = new JsonTextReader(tr))
            {
                var rootObjectResponse = new JsonSerializer    
().Deserialize<JsonDOTNetHelperClasses.RootObject>(jr);    
                var firstplace = rootObjectResponse.results.items.First();
                return HtmlUtilities.ConvertToText(firstplace.vicinity);
                // NOTE: There is also a title (such as "Donut Shop", "Fire stations", etc.?) and type (such as "residence" or "business", etc.?)
            }
        }

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...但是在GetAddressForCoordinates()的这一行:

        var firstplace = rootObjectResponse.results.items.First();

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...我得到这个错误信息:"*System.InvalidOperationException未被用户代码处理HResult = -2146233079 Message = Sequence不包含元素Source = System.Core StackTrace:at System.Linq.Enumerable.First [TSource](IEnumerable` 1来源)在C的SpaceOverlays.SOs_Classes.SOs_Utils.d__12.MoveNext():**"

hereNetResponse的值是:

{"results":{"items":[]},"search":{"context":{"location":{"position":[38.804967,-90.113183],"address":
{"postalCode":"62048","city":"Hartford","stateCode":"IL","county":"Madison","countryCode":"USA","country":"
USA","text":"Hartford IL 62048
USA"}},"type":"urn:nlp-types:place","href":"http://demo.places.nlp.nokia.com/places/v1/places/loc-
dmVyc2lvbj0xO3RpdGxlPUhhcnRmb3JkO2xhdD0zOC44MDQ5Njc7bG9uPS05MC4xMTMxODM7Y2l0eT1IY
XJ0Zm9yZDtwb3N0YWxDb2RlPTYyMDQ4O2NvdW50cnk9VVNBO3N0YXRlQ29kZT1JTDtjb3VudHk9TWFka
XNvbjtjYXRlZ29yeUlkPWNpdHktdG93bi12aWxsYWdl;context=Zmxvdy1pZD02YmUzZDM4Yi0wNGVhLTUyM
jgtOWZmNy1kNWNkZGM0ODI5OThfMTM1NzQyMDI1NTg1M18wXzE2MA?
app_id=F6zpNc3TjnkiCLwl_Xmh&app_code=QoAM_5BaVDZvkE2jRvc0mw"}}}

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...所以看来里面有有效的信息,比如应该返回"哈特福德,伊利诺伊"

无论如何,空白的返回值不应该抛出异常,我想......

Answer 1

Pet*_*hie 14

您要问的只是"地理编码".如果你想特别使用Nominatim,他们称之为"搜索".这在某种程度上是地址验证; 但"验证"的一部分包括坐标(边界框,纬度/经度等;取决于搜索的内容和结果的类型).关于结果有很多细节,太多了,不能简单地在这里发布; 但是这个细节可以在这里找到:http://wiki.openstreetmap.org/wiki/Nominatim#Search(包括考试).

您必须解析结果(XML,JSON或HTML)以获取您感兴趣的字段.

更新1:

至于如何处理实际值:它取决于.如果要查看表单中的坐标,只需将lat和long字符串放入单独的控件中即可.如果你想把它放在一个控件中,可以使用string.Format("{0}, {1}", latString, longString).如果要对Windows应用商店应用使用各种方法/类型的坐标,则可能需要使用Microsoft.Maps.MapControl.Location该类.例如:

  Double latNumber;
  Double longNumber;
  if(false == Double.TryParse(latString, out latNumber)) throw new InvalidOperationException();
  if(false == Double.TryParse(longString, out longNumber)) throw new InvalidOperationException();
  var location = new Location(latNumber, longNumber);

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上述假设你已经提取lat和长从响应,并把他们在latString,longString分别.

某些接口可能需要lat/long作为单独的double值,在这种情况下只需使用latNumber和longNumber以上.

除此之外,它实际上特别取决于您要使用的接口.但是,上面应该足以让你使用大多数接口.

更新2:

如果问题不是"如何获取坐标"而是"如何解析json对象",那么我建议使用JSon.Net来获取json结果中的lat/long字符串.例如:

    var httpClient = new HttpClient();
    var httpResult = await httpClient.GetAsync(
        "http://nominatim.openstreetmap.org/search?q=135+pilkington+avenue,+birmingham&format=json&polygon=1&addressdetails=1");

    var result = await httpResult.Content.ReadAsStringAsync();
    var r = (JArray) JsonConvert.DeserializeObject(result);
    var latString = ((JValue) r[0]["lat"]).Value as string;
    var longString = ((JValue)r[0]["lon"]).Value as string;

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...看看上面怎么做latString和longString

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