如何使用 C# 從 SQL 服務器檢索點？

Question

i'm using spacial data type on my sql server and I want to retrieve this data from my sql server to my Entity Class, it returns the error ".NET number values such as positive and negative infinity cannot be written as valid JSON." 當我嘗試從我的數據庫中獲取值時。

坐標是我想從我的數據庫返回的值。 我正在使用來自 NetTopologySuite 的地理數據類型，並且正在從我的實體 Class 中插入一個點。

我正在使用的 class 代碼：

public class Address : EntityBase<int>
{
    public Address () {}

    public Address(string district, string street, int number, string complement, string zipCode, 
    string cityDescription, string stateDescription, Point coordinates, int countryId, byte stateId, int cityId)
    {
        District = district;
        Street = street;
        Number = number;
        Complement = complement;
        ZipCode = zipCode;
        CityDescription = cityDescription;
        StateDescription = stateDescription;
        Coordinates = coordinates;
        CountryId = countryId;
        StateId = stateId;
        CityId = cityId;
    }

    public string District { get; set; }
    public string Street { get; set; }
    public int Number { get; set; }
    public string Complement { get; set; }
    public string ZipCode { get; set; }
    public string CityDescription { get; set; }
    public string StateDescription { get; set; }
    public virtual Point Coordinates { get; set; }
    public int CountryId { get; set; }
    public byte StateId { get; set; }
    public int CityId { get; set; }
    public int? CustomerId { get; set; }
    public int? StoreId { get; set; }
    public int? ProfessionalId { get; set; }
}

以及表的 sql 服務器代碼：

CREATE TABLE Address(
    Id INT PRIMARY KEY IDENTITY(1, 1),
    District VARCHAR(50) NOT NULL, -- Bairro
    Street VARCHAR(100) NOT NULL, -- Rua
    --Description VARCHAR(100) NOT NULL,
    Number INT,
    Complement VARCHAR(100),
    ZipCode VARCHAR(20) NOT NULL,
    CityDescription VARCHAR(100),
    StateDescription VARCHAR(100),
    Coordinates GEOGRAPHY,
    
    CountryId INT FOREIGN KEY REFERENCES Country(Id) NOT NULL,
    StateId TINYINT FOREIGN KEY REFERENCES State(Id),
    CityId INT FOREIGN KEY REFERENCES City(Id),
    CustomerId INT FOREIGN KEY REFERENCES Customer(Id),
    StoreId INT FOREIGN KEY REFERENCES Store(Id),
    ProfessionalId INT FOREIGN KEY REFERENCES Professional(Id),

    INDEX IndexAddressCountryId NONCLUSTERED (CountryId),
    INDEX IndexAddressStateId NONCLUSTERED (StateId),
    INDEX IndexAddressCityId NONCLUSTERED (CityId),
    INDEX IndexAddressCustomerId NONCLUSTERED (CustomerId),
    INDEX IndexAddressStoreId NONCLUSTERED (StoreId),
    INDEX IndexAddressProfessionalId NONCLUSTERED (ProfessionalId)
)

有什么方法可以從中檢索點值嗎？ 像 function OnModelCreating上的配置還是其他什么？ 我只能從中檢索經度和緯度嗎？

我是空間數據的新手，所以我不太了解它。 提前感謝您幫助我:)

---

編輯1：

這是我得到的異常錯誤：

實體庫：

namespace CodenApp.Framework.Core.Entities
{
    public abstract class EntityBase<T>
    {
        public T Id { get; set; }
    }
}

還有我用來處理異常的兩個函數：

public async Task InvokeAsync(HttpContext httpContext)
{
    try
    {
        await _next(httpContext);
    }
    catch (Exception ex)
    {
        _logger.LogError(ex, ex.ToString());                
        await HandleExceptionAsync(httpContext, ex);
    }
}


protected override Task HandleExceptionAsync(HttpContext context, Exception exception)
{
    if(context == null)
        throw new ArgumentNullException(nameof(context));
    if(exception == null)
        throw new ArgumentNullException(nameof(exception));

    context.Response.ContentType = "application/json";
    context.Response.StatusCode = (int)HttpStatusCode.BadRequest;
    return context.Response.WriteAsync(ApiCatch.Log(exception).ToString());
}

我用來轉換為 Point 的代碼：

public static Point ConvertToGeometry(string coordinatesAux)
{
    if(coordinatesAux == null)
        throw new NullReferenceException();

    NumberFormatInfo formatProvider = new NumberFormatInfo();
    formatProvider.NumberGroupSeparator = ".";
            
    var currentLocation = new Point(new Coordinate(
                                        Convert.ToDouble(coordinatesAux.Substring(0, coordinatesAux.IndexOf(",")), formatProvider), 
                                        Convert.ToDouble(coordinatesAux.Substring(coordinatesAux.IndexOf(",") + 1), formatProvider))) { SRID = 4326 };
            
    return currentLocation;                                                            
}

Answer 1

As Panagiotis Kanavos said in the comments, to solve this situation I needed to add the NuGet package NetTopologySuite.IO.GeoJSON4STJ and change the file Startup.cs to recieve the changes given by the package.

ConfigureServices function 內的 Startup.cs 更改：

services.AddControllers(options =>
{
    options.ModelMetadataDetailsProviders.Add(new SuppressChildValidationMetadataProvider(typeof(Point))); 
    options.ModelMetadataDetailsProviders.Add(new SuppressChildValidationMetadataProvider(typeof(Coordinate))); 
    options.ModelMetadataDetailsProviders.Add(new SuppressChildValidationMetadataProvider(typeof(LineString))); 
    options.ModelMetadataDetailsProviders.Add(new SuppressChildValidationMetadataProvider(typeof(MultiLineString))); 
});

services.AddControllers().AddJsonOptions(options =>
{
    var geoJsonConverterFactory = new GeoJsonConverterFactory();
    options.JsonSerializerOptions.Converters.Add(geoJsonConverterFactory);
});

services.AddSingleton(NtsGeometryServices.Instance);

並將 IOptions 添加到我的 Controller 中：

private readonly IOptions<JsonOptions> _jsonOptions;
public StoreController(IAsyncRepository<Store, int> repository, StoreHandler handler, IOptions<JsonOptions> jsonOptions) : base(repository)
{
    _handler = handler;
    _orders = new Expression<Func<Store, object>>[] { x => x.Id };
    _jsonOptions = jsonOptions;
}

所以在這個wiki中教授。

Answer 2

您可以簡單地 JSON 忽略導致問題的屬性並創建以可接受的格式顯示值的屬性。 這是我的做法：

public class LocationResult
{
    public Guid id { get; set; }
    [JsonIgnore]
    public GeoCoordinate geo { get; set; }
    public double latitude
    {
        get
        {
            return geo.Latitude;
        }
    }
    public double longitude
    {
        get
        {
            return geo.Longitude;
        }
    }
}

請注意：我知道這是針對與上面列出的不同的 object 類型，但是該方法仍然適用於調整后的屬性。