c# - 嵌套的 foreach 循环读取 .xml 并将对象写入列表

Question

我有一个在 form_load 事件上执行的方法，似乎可以正常工作，省略一行。

private int ReadInPeople()
{
    XmlNodeList nodeList = m_xmlDoc.DocumentElement.ChildNodes;
    foreach (XmlNode PersonNode in nodeList)
    {
        Employee ccontact = new Employee();
        foreach (XmlNode PersonTag in PersonNode.ChildNodes)
        {
            switch (PersonTag.Name)
            {
                case "Employee":
                    ccontact.EmployeeNumber = PersonTag.FirstChild.Value;
                    break;
                case "FirstName":
                    ccontact.FirstName = PersonTag.FirstChild.Value;
                    break;
                case "LastName":
                    ccontact.LastName = PersonTag.FirstChild.Value;
                    break;
                default:
                    break;
            }
        }
        this.AddContact(ccontact);
    }
    return nodeList.Count;
}

AddContact 方法将 Employee 对象添加到静态列表；但是，行:

this.AddContact(ccontact);

没有被执行。

XML 文件示例:

<?xml version="1.0" encoding="utf-8"?>
<people>
  <person>
    <Employee>123456789</Employee>
    <FirstName>John</FirstName>
    <LastName>Smith</LastName>
  </person>
  <person>
    <Employee>987654321</Employee>
    <FirstName>Ellen</FirstName>
    <LastName>Wayne</LastName>
  </person>
</people>

我试过设置断点和调试，果然，这一行被完全跳过了，就好像它根本不存在一样。

根据 Alan 的建议，我更改了 PersonTag.FirstChild.Value，因为它试图引用不存在的 ChildNode。

更新后的工作方法:

private int ReadInPeople()
{
    XmlNodeList nodeList = m_xmlDoc.DocumentElement.ChildNodes;
    foreach (XmlNode PersonNode in nodeList)
    {
        Employee ccontact = new Employee();
        foreach (XmlNode PersonTag in PersonNode.ChildNodes)
        {
            switch (PersonTag.Name)
            {
                case "Employee":
                    ccontact.EmployeeNumber = PersonTag.InnerText;
                    break;
                case "FirstName":
                    ccontact.FirstName = PersonTag.InnerText;
                    break;
                case "LastName":
                    ccontact.LastName = PersonTag.InnerText;
                    break;
                default:
                    break;
            }
        }
        this.AddContact(ccontact);
    }
    return nodeList.Count;
}

Answer 1

您可以使用 LINQ to XML 轻松解析您的 xml:

XDocument xdoc = XDocument.Load(path_to_xml);   
var employees = xdoc.Descendants("person")
                    .Select(p => new Employee()
                    {
                        FirstName = (string)p.Element("FirstName"),
                        LastName = (string)p.Element("LastName"),
                        EmployeeNumber = (long)p.Element("Employee")
                    });

foreach (var ccontact in employees)
    this.AddContact(ccontact);

---

XmlDocument解决方案:

XmlNodeList nodeList = m_xmlDoc.DocumentElement.SelectNodes("person");
foreach (XmlNode PersonNode in nodeList)
{
    Employee ccontact = new Employee();
    ccontact.LastName = PersonNode["LastName"].InnerText;
    ccontact.FirstName = PersonNode["FirstName"].InnerText;
    ccontact.EmployeeNumber = PersonNode["Employee"].InnerText;
    this.AddContact(ccontact);
}