我正在尝试使用C#从XML文件中提取特定节点。我想从此XML中仅提取userID(123456789)
<Response Destination="https://saml.qc.xxxx.com/sp/ACS.saml2" IssueInstant="2011-02-14T20:39:00.328Z" ID="iRTyBb7E9OLitdGZT1RYRSJNX85" Version="2.0" xmlns="urn:oasis:names:tc:SAML:2.0:protocol" xmlns:saml="urn:oasis:names:tc:SAML:2.0:assertion" xmlns:ds="http://www.w3.org/2000/09/xmldsig#" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<saml:Issuer>some.client.com:sso
</saml:Issuer>
<Status>
<StatusCode Value="urn:oasis:names:tc:SAML:2.0:status:Success" />
</Status>
<saml:Assertion Version="2.0" IssueInstant="2011-02-14T20:39:00.328Z" ID="YJtYu1HoChn0nrORzDSkVGOE8RD">
<saml:Issuer>some.client.com:sso</saml:Issuer>
<saml:Subject>
<saml:NameID Format="urn:oasis:names:tc:SAML:1.1:nameid-format:unspecified">123456789</saml:NameID>
<saml:SubjectConfirmation Method="urn:oasis:names:tc:SAML:2.0:cm:bearer">
<saml:SubjectConfirmationData NotOnOrAfter="2011-02-14T20:40:00.328Z" Recipient="https://saml.qc.xxxx.com/sp/ACS.saml2" />
</saml:SubjectConfirmation>
</saml:Subject>
<saml:Conditions NotOnOrAfter="2011-02-14T20:40:00.328Z" NotBefore="2011-02-14T20:38:00.328Z">
<saml:AudienceRestriction>
<saml:Audience>saml.qc.xxxx.com:saml2.0</saml:Audience>
</saml:AudienceRestriction>
</saml:Conditions>
<saml:AuthnStatement AuthnInstant="2011-02-14T20:39:00.328Z" SessionIndex="YJtYu1HoChn0nrORzDSkVGOE8RD">
<saml:AuthnContext>
<saml:AuthnContextClassRef>urn:oasis:names:tc:SAML:2.0:ac:classes:Pwd</saml:AuthnContextClassRef>
</saml:AuthnContext>
</saml:AuthnStatement>
<saml:AttributeStatement xmlns:xs="http://www.w3.org/2001/XMLSchema">
<saml:Attribute NameFormat="urn:oasis:names:tc:SAML:2.0:attrname-format:basic" Name="clientId">
<saml:AttributeValue xsi:type="xs:string">99999</saml:AttributeValue>
</saml:Attribute>
</saml:AttributeStatement>
</saml:Assertion>
</Response>
这是我的代码:
public static void ExtractUserID(XmlDocument Doc)
{
XmlElement docRoot = Doc.DocumentElement;
XmlNodeList idNode = Doc.GetElementsByTagName("saml:NameID");
Console.WriteLine("UserID: " + idNode);
Console.ReadLine();
}
输出:
UserID: System.Xml.XmlElementList
我想得到这样的东西:
User ID: 123456789
答案 0 :(得分:0)
使用xml linq:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;
namespace ConsoleApplication120
{
class Program
{
const string FILENAME = @"c:\temp\test.xml";
static void Main(string[] args)
{
XDocument doc = XDocument.Load(FILENAME);
string id = (string)doc.Descendants().Where(x => x.Name.LocalName == "NameID").FirstOrDefault();
}
}
}
答案 1 :(得分:0)
您可以同时使用Linq和XDocument。
将您的文件定义为XDoc,我刚刚将其命名为dDoc
XDocument Doc = XDocument.Parse(your file/string);
然后定义一个列表,以保存saml字符串中的所有节点,并使用链接过滤NameID
List<XElement> userIDs = (from element in Doc.Descendants()
.Where(x => x.Name.LocalName.Contains("NameID"))
select element).ToList();
然后将所有结果列出到控制台
foreach (XElement element in userIDs)
{
Console.WriteLine(element.Value);
}