我正在构建一个必须返回json响应的服务:
{“内容”:{“API”:“某些API”,“DisplayVersion”:“1.0”,“版本”:1},“状态”:“确定”}
我发现很难按照我想要的方式格式化它。我正以这种方式得到回应:
{“API”:“某些API”,“DisplayVersion”:“1.0”,“版本”:“1”}
以下是我的相同代码:
[WebInvoke(Method = "GET", BodyStyle =
WebMessageBodyStyle.WrappedRequest, RequestFormat =
WebMessageFormat.Json, ResponseFormat = WebMessageFormat.Json,
UriTemplate = "")]
public Info GetSystemInfo()
{
return new Info()
{
API = "Some API",
DisplayVersion = "1.0",
Version = "1"
};
}
[DataContract]
public class Info
{
[DataMember]
public string API { get; set; }
[DataMember]
public string DisplayVersion { get; set; }
[DataMember]
public string Version { get; set; }
}
我试着像这样构建我的类:
[DataContract]
public class Info
{
[DataMember]
public class Content
{
[DataMember]
public string API { get; set; }
[DataMember]
public string DisplayVersion { get; set; }
[DataMember]
public string Version { get; set; }
}
[DataMember]
public string Status { get; set; }
}
我应该怎么做呢?
答案 0 :(得分:2)
尝试按照以下方式构建类:
[DataContract]
public class Info
{
[DataMember]
public Content Content { get; set; }
[DataMember]
public string Status { get; set; }
}
public class Content
{
[DataMember]
public string API { get; set; }
[DataMember]
public string DisplayVersion { get; set; }
[DataMember]
public int Version { get; set; }
}
编辑:使用信息:
Info info = new Info()
{
Status = "OK",
Content = new Content() { API = "Some Api", DisplayVersion = "1.0", Version = 1 }
};