Question

我有这个XML输入：

<Response xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\"><Status>Ok</Status></Response>

我试图通过以下方式将其转换为对象：

XmlDocument xmlNode = new XmlDocument();
xmlNode.LoadXml(responseContent);
string jsonNode = JsonConvert.SerializeXmlNode(xmlNode);
var responseModel = JsonConvert.DeserializeObject<NotificationResponse>(jsonNode);

其中responseContent是上面的XML字符串。但是，而不是

{Status: "Ok"}

我得到了：

{Status: null}

知道为什么以及如何纠正它？

Answer 1

您需要省略RootObject

XmlDocument xmlNode = new XmlDocument();
xmlNode.LoadXml("<Response xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\"><Status>Ok</Status></Response>");
string jsonNode = JsonConvert.SerializeXmlNode(xmlNode, Formatting.Indented, true);
Console.WriteLine(jsonNode);

var responseModel = JsonConvert.DeserializeObject<NotificationResponse>(jsonNode);
Console.WriteLine(responseModel.Status);

Answer 2

更改为string jsonNode = JsonConvert.SerializeXmlNode(xmlNode.DocumentElement.SelectSingleNode("Status"));

        XmlDocument xmlNode = new XmlDocument();
        xmlNode.LoadXml("<Response xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\"><Status>Ok</Status></Response>");
        string jsonNode = JsonConvert.SerializeXmlNode(xmlNode.DocumentElement.SelectSingleNode("Status"));

        var responseModel = JsonConvert.DeserializeObject<NotificationResponse>(jsonNode);

Xml无法正确转换为对象

2 个答案: