[英]RestSharp: Deserialize Xml to c# Object return null
我有一些api的xml:
<auditypes>
<auditype code="a" description="aaa"/>
<auditype code="b" description="bbb"/>
<auditype code="c" description="ccc"/>
<auditype code="d" description="ddd"/>
<auditype code="e" description="eee"/>
</auditypes>
和映射为C#类中的对象:
public class auditypes
{
public List<auditype> auditype { get; set; }
}
public class auditype
{
public string code { get; set; }
public string description { get; set; }
}
我用这个函数来称呼它:
public List<auditypes> Execute<auditypes>(RestRequest request) where auditypes : new()
{
var client = new RestClient();
client.BaseUrl = "https://www.myurl.com/auditypes";
var response = client.Execute<auditypes>(request);
return response.Data as List<auditypes>;
}
public List<auditypes> GetCall()
{
var request = new RestRequest();
request.RequestFormat = DataFormat.Xml;
request.RootElement = "auditype";
return Execute<auditypes>(request);
}
但是它总是返回null,有人知道为什么会这样吗?
传递给Execute<T>
的通用参数是应由RestSharp
库反序列化的类型。 这意味着您的response.Data
属性已经是T
类型,在您的情况下是auditypes
。 但是,当您return
,尝试将其List<auditypes>
不存在此类List<auditypes>
转换的List<auditypes>
。
另外,不需要类型约束,因为您的方法不是通用的,因为它接受显式类型。
重构您的方法:
public auditypes Execute<auditypes>(RestRequest request)
{
var client = new RestClient();
client.BaseUrl = "https://www.myurl.com/auditypes";
var response = client.Execute<auditypes>(request);
return response.Data;
}
最后,它对我有用:)
public auditypes Execute<auditypes>(RestRequest request) where auditypes : new()
{
var client = new RestClient();
client.BaseUrl = "https://www.myurl.com/auditypes";
var response = client.Execute<auditypes>(request).Data;
return response;
}
public auditypes GetCall()
{
var request = new RestRequest();
request.RequestFormat = DataFormat.Xml;
return Execute<auditypes>(request);
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.