读取 XML 时未将对象引用设置为对象的实例

Question

这是我需要处理的 XML 文件：

<?xml version="1.0" encoding="UTF-8"?>
<shipment-info xmlns="http://www.canadapost.ca/ws/shipment-v8">
    <shipment-id>11111111</shipment-id>
    <shipment-status>created</shipment-status>
    <tracking-pin>123456789012</tracking-pin>
    <links>
        <link rel="self" href="https://xxx" media-type="application/vnd.cpc.shipment-v8+xml"/>
        <link rel="details" href="https://xxx" media-type="application/vnd.cpc.shipment-v8+xml"/>
        <link rel="group" href="https://xxx" media-type="application/vnd.cpc.shipment-v8+xml"/>
        <link rel="price" href="https://xxx" media-type="application/vnd.cpc.shipment-v8+xml"/>
        <link rel="label" href="https://xxx" media-type="application/pdf" index="0"/>
    </links>
</shipment-info>

我想获取跟踪引脚值，这是执行此操作的代码：

// xml text
string xml = (xml source above)

// load xml
XmlDocument doc = new XmlDocument();
doc.LoadXml(xml);

// getting tracking pin
XmlNode node = doc.SelectSingleNode("/shipment-info"); // problem start here -> node return null

Console.WriteLine(node["tracking-pin"].InnerText);

// getting self and label links
node = doc.SelectSingleNode("/shipment-info/links");
foreach (XmlNode child in node)
{
    if (child.Attributes["rel"].Value == "self")
        Console.WriteLine(child.Attributes["href"].Value);
    else if (child.Attributes["rel"].Value == "label")
        Console.WriteLine(child.Attributes["href"].Value);
}

Console.ReadLine();

但是，对于选择“/shipment-info”，它返回一个空值，我不知道为什么会发生这种情况。如何处理？

Answer 1

您有一个需要考虑的命名空间：

// load xml
XmlDocument doc = new XmlDocument();
doc.LoadXml(xml);

System.Xml.XmlNamespaceManager xmlnsManager = new System.Xml.XmlNamespaceManager(doc.NameTable);

//Add the namespaces used in books.xml to the XmlNamespaceManager.
xmlnsManager.AddNamespace("veeeight", "http://www.canadapost.ca/ws/shipment-v8");


// getting tracking pin
XmlNode node = doc.SelectSingleNode("//veeeight:shipment-info", xmlnsManager); // problem start here -> node return null

见：

https://support.microsoft.com/en-us/kb/318545

Answer 2

您遇到了命名空间问题。试试这个 XML Linq 解决方案

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

namespace ConsoleApplication1
{
    class Program
    {
        const string FILENAME = @"c:\temp\test.xml";
        static void Main(string[] args)
        {
            XDocument doc = XDocument.Load(FILENAME);
            XNamespace ns = ((XElement)(doc.FirstNode)).Name.Namespace;
            string tracking_pin = doc.Descendants(ns + "tracking-pin").FirstOrDefault().Value;
        }
    }
}

Answer 3

将其添加到您的代码中

XmlNode root = doc.DocumentElement;