c# - 如何在类属性中分配值类

Question

<分区>

Answer 1

     public void ConvertJson()
     {
         BookingInformation objbooking = new BookingInformation();
         objbooking.booking_id = "33";
         objbooking.to_location.comment= "Saloon black";
         objbooking.to_location.address.country= "UK";
         var json = JsonConvert.SerializeObject(objbooking);
     }

您还没有对 ToLocation 实例的引用。更改为:

     public void ConvertJson()
     {
         BookingInformation objbooking = new BookingInformation();
         objbooking.to_location = new ToLocation();
         objbooking.to_location.address = new Address2();
         objbooking.booking_id = "33";
         objbooking.to_location.comment= "Saloon black";
         objbooking.to_location.address.country= "UK";
         var json = JsonConvert.SerializeObject(objbooking);
     }

根据您的编辑，您还没有 Address2 的实例，我也更新了上面的内容以说明这一点。

另请通读以下回复:What is a NullReferenceException, and how do I fix it?有关 NullReferenceExceptions 如何/为何发生的更多信息。

另一种解决这个问题的方法是从父类的构造开始，这样您就不必记住更新需要它的对象的属性，如下所示:

public class BookingInformation
{
    public string booking_id { get; set; }
    public ToLocation to_location { get; set; }
    public string notes { get; set; }

    public BookingInformation()
    {
        to_location = new ToLocation();
    }
}

public class ToLocation
{
    public double latitude { get; set; }
    public double longitude { get; set; }
    public Address2 address { get; set; }
    public object comment { get; set; }
    public object airport { get; set; }

    public ToLocation()
    {
        address = new Address2();
    }
}

这将确保当您创建 BookingInformation 的新实例时，ToLocation 又会在构造时创建 Address2 实例。