Question

请帮助我解决问题，如何读取属性和嵌套XML。我不明白这个原则，我想在真实的例子上讲这个。

来源xml：

let appSettings = ConfigurationManager.AppSettings
let value = appSettings.Item(key)

我想读取用户uuid属性以及要追加到DataTable的所有节点。但我不知道如何读取属性和嵌套的xml（错误：“对象引用未设置为对象的实例。”）

我的代码：

<?xml version="1.0" encoding="utf-8"?>
<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/">
  <soapenv:Body>
    <ns:listUserResponse xmlns:ns="http://www.cisco.com/AXL/API/11.5">
      <return>
        <user uuid="{A00B47CC-0C4F-40E7-1E25-22E54FBA4596}">
          <firstName>Name1</firstName>
          <middleName />
          <lastName>Surname1</lastName>
          <primaryExtension>
            <pattern>201</pattern>
          </primaryExtension>
        </user>
        <user uuid="{834E86FA-FDC2-21FE-1E1E-FD164AF23B44}">
          <firstName>Name2</firstName>
          <middleName />
          <lastName>Surname2</lastName>
          <primaryExtension>
            <pattern>202</pattern>
          </primaryExtension>
        </user>
        <user uuid="{798E04E7-0E10-0278-6FAD-EA6A85EF2B2E}">
          <firstName>Name3</firstName>
          <middleName />
          <lastName>Surname3</lastName>
          <primaryExtension>
            <pattern>203</pattern>
          </primaryExtension>
        </user>
      </return>
    </ns:listUserResponse>
  </soapenv:Body>
</soapenv:Envelope>

谢谢你，最诚挚的问候，彼得

Answer 1

你可以使用这样的东西。

foreach (XmlNode xn in xnList)
    {
        string uuid = (xn as XmlElement).GetAttribute("uuid");
        string firstName = xn.SelectSingleNode("./firstName").InnerText;
        string middleName= xn.SelectSingleNode("./middleName").InnerText;
        string lastName = xn.SelectSingleNode("./lastName").InnerText;
        string primaryExtensionPattern = xn.SelectSingleNode("./primaryExtensionPattern /pattern").InnerText;

        ...
    }

或者您可以考虑将Linq用于XML并使用XElement。对我来说，似乎更容易解析XML

Answer 2

试试xml linq

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Data;
using System.Xml;
using System.Xml.Linq;


namespace ConsoleApplication1
{
    class Program
    {
        const string FILENAME = @"c:\temp\test.xml";
        static void Main(string[] args)
        {
            DataTable datatableListUser = new DataTable();
            datatableListUser.Clear();
            datatableListUser.Columns.Add("uuid");
            datatableListUser.Columns.Add("firstName", typeof(string));
            datatableListUser.Columns.Add("middleName", typeof(string));
            datatableListUser.Columns.Add("lastName", typeof(string));
            datatableListUser.Columns.Add("primaryExtensionPattern", typeof(int));

            XDocument doc = XDocument.Load(FILENAME);

            foreach (XElement user in doc.Descendants("user"))
            {
                datatableListUser.Rows.Add(new object[] {
                   (string)user.Attribute("uuid"),
                   (string)user.Element("firstName"),
                   (string)user.Element("middleName"),
                   (string)user.Element("lastName"),
                   (int)user.Descendants("pattern").FirstOrDefault()
                });
            }

        }
    }
}

如何读取属性和嵌套XML

2 个答案: