反序列化XML，ASP时缺少根元素 [英] Root element is missing when deserialize XML , ASP

问题描述

我从服务中获取xml文件，但是当我想要

i get xml file from service , but i have problem when i want

Deserialize 

来对象时，我有问题，总是显示

to object ,, always show me

root element is missing 

xml

xml

<?xml version="1.0" encoding="utf-8"?>
<ArrayOfResultInfo xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://tempuri.org/">
<ResultInfo>
<PartNumber>KRPA-4007</PartNumber>
<Manufacturer>test</Manufacturer>

</ResultInfo>
</ArrayOfResultInfo>

code

code

string xml = null;
     WebRequest req = WebRequest.Create(link);
     req.Credentials = CredentialCache.DefaultCredentials;
     WebResponse res = req.GetResponse();
     Stream dataStream = res.GetResponseStream();
     StreamReader reader = new StreamReader(dataStream);
     xml = reader.ReadToEnd();
     XmlSerializer serializer = new XmlSerializer(typeof(ArrayOfResultInfo));
     var result = (ArrayOfResultInfo)serializer.Deserialize(reader);

我的尝试：



string xml = null;

WebRequest req = WebRequest.Create（link）;

req.Credentials = CredentialCache.DefaultCredentials;

WebResponse res = req.GetResponse（）;

流dataStream = res.GetResponseStream（）;

StreamReader reader = new StreamReader（dataStream）;

xml = reader.ReadToEnd（）;

XmlSerializer serializer = new XmlSerializer（typeof（ArrayOfResultInfo））;

var result =（ArrayOfResultInfo）serializer.Deserialize（reader ）;

What I have tried:

string xml = null;
WebRequest req = WebRequest.Create(link);
req.Credentials = CredentialCache.DefaultCredentials;
WebResponse res = req.GetResponse();
Stream dataStream = res.GetResponseStream();
StreamReader reader = new StreamReader(dataStream);
xml = reader.ReadToEnd();
XmlSerializer serializer = new XmlSerializer(typeof(ArrayOfResultInfo));
var result = (ArrayOfResultInfo)serializer.Deserialize(reader);

推荐答案

这可以帮到你：将XML转换为C＃类 [ ^ ]

This may help you: Convert XML to C# Classes[^]

以下代码适用于我的xml：

Hi,

The following code worked for me with your xml:

string xml = "<?xml version=\"1.0\" encoding=\"utf-8\"?><ArrayOfResultInfo xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns=\"http://tempuri.org/\"><ResultInfo><PartNumber>KRPA - 4007 </PartNumber><Manufacturer> test </Manufacturer></ResultInfo></ArrayOfResultInfo>";



           var serializer = new XmlSerializer(typeof(ArrayOfResultInfo));
           ArrayOfResultInfo result;

           using (TextReader reader = new StringReader(xml))
           {
               result = (ArrayOfResultInfo)serializer.Deserialize(reader);
           }

           Console.WriteLine(result.ResultInfo.Manufacturer);
           Console.WriteLine(result.ResultInfo.PartNumber);

           Console.ReadLine();

xml如下：

The classes for xml are as follows:

[XmlRoot(ElementName = "ResultInfo", Namespace = "http://tempuri.org/")]
public class ResultInfo
{
    [XmlElement(ElementName = "PartNumber", Namespace = "http://tempuri.org/")]
    public string PartNumber { get; set; }
    [XmlElement(ElementName = "Manufacturer", Namespace = "http://tempuri.org/")]
    public string Manufacturer { get; set; }
}

[XmlRoot(ElementName = "ArrayOfResultInfo", Namespace = "http://tempuri.org/")]
public class ArrayOfResultInfo
{
    [XmlElement(ElementName = "ResultInfo", Namespace = "http://tempuri.org/")]
    public ResultInfo ResultInfo { get; set; }
    [XmlAttribute(AttributeName = "xsd", Namespace = "http://www.w3.org/2000/xmlns/")]
    public string Xsd { get; set; }
    [XmlAttribute(AttributeName = "xsi", Namespace = "http://www.w3.org/2000/xmlns/")]
    public string Xsi { get; set; }
    [XmlAttribute(AttributeName = "xmlns")]
    public string Xmlns { get; set; }
}

希望这会有所帮助！

Hope this helps !

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