0

我一直在尝试如何用 XML 制作数据库。我已成功写入数据，如下所示： 

<Employees>
    <Worker>
        <ID>1</ID>
        <FirstName>Ilan</FirstName>
        <LastName>Berlinbluv</LastName>
        <Salary>5000</Salary>
    </Worker>
</Employees>

问题是当我尝试使用此代码阅读时： 

    string writePath = Environment.ExpandEnvironmentVariables("%USERPROFILE%") + @"\Desktop";
    string writeFile = writePath + @"\Employees.xml";

    using (XmlReader read = XmlReader.Create(writeFile))
    {
        while (read.Read())
        {
            if (read.IsStartElement())
            {
                Console.WriteLine("DEBUG: read.Name = {0}", read.Name);
                Console.WriteLine("DEBUG: read.Value = {0}", read.Value);
                switch (read.Name)
                {
                    case "Employees":
                        Console.WriteLine("Start <Employees> master element");
                        break;
                    case "Employee":
                        Console.WriteLine("Start <Employee> element");
                        break;
                    case "Worker":
                        Console.WriteLine("Start <Worker> element");
                        break;
                    case "ID":
                        Console.WriteLine("Start reading <ID> element");
                        Console.WriteLine("Contains: " + read.Value.Trim());
                        break;
                    case "FirstName":
                        Console.WriteLine("Start reading <FirstName> element");
                        Console.WriteLine("Contains: " + read.Value.Trim());
                        break;
                    case "LastName":
                        Console.WriteLine("Start reading <LastName> element");
                        Console.WriteLine("Contains: " + read.Value.Trim());
                        break;
                    case "Salary":
                        Console.WriteLine("Start reading <Salary> element");
                        Console.WriteLine("Contains: " + read.Value.Trim());
                        break;
                }
            }
            Console.ReadKey();
        }
    }

它没有正确读取值，每当它说Start reading <Salary> elementthenContains:时，它不显示任何值，但应该有一个值：5000。它是语法错误吗，我需要它是这样的： 

<ID>
1
</ID>

我一直在做dotnetperls 教程，但无济于事。

4

6 回答 6

0

您需要使用 read.InnerText 而不是 read.Value。

一旦您了解了所有这些是如何工作的，我建议您使用内置的 XML 处理函数。你会省去很多麻烦。

于 2013-10-31T10:58:12.053 回答
0

您可以使用XSD 工具来使用序列化程序。首先创建一个 XML 模式，然后从 XML 模式创建一个类。然后您可以将 XML 直接读取到对象中：

        /// <summary>
    /// 
    /// </summary>
    /// <typeparam name="T"></typeparam>
    /// <param name="xml"></param>
    /// <exception cref="InvalidOperationException"></exception>
    /// <returns></returns>
    public static T Deserialize<T>(XmlNode xml)
    {
        // Assuming xml is an XML document containing a serialized object.
        XmlNodeReader reader = new XmlNodeReader(xml);

        // When we get the xml, it is usually a sub element that can have a different name, than the type name. Therefore look for the name
        XmlSerializer ser = new XmlSerializer(typeof(T));
        object obj = ser.Deserialize(reader);

        // Then you just need to cast obj into whatever type it is eg:
        return (T)obj;
    }
    /// <summary>
    /// Serializes without removing namespace and using the specified encoding
    /// </summary>
    /// <typeparam name="T"></typeparam>
    /// <param name="obj"></param>
    /// <param name="encoding"></param>
    /// <returns></returns>
    /// <exception cref="InvalidOperationException">When the object can not be serialized to xml</exception>
    public static XmlDocument Serialize<T>(T obj, Encoding encoding)
    {
        XmlSerializer ser = GetSerializer(obj.GetType());
        using (MemoryStream stream = new MemoryStream())
        using (XmlTextWriter writer = new XmlTextWriter(stream, encoding))
        {
            ser.Serialize(writer, obj);
            XmlDocument doc = new XmlDocument();
            writer.Flush();
            stream.Position = 0;
            doc.Load(stream);
            return doc;
        }
    }

您可能希望在创建类之前编辑 XML 模式以确保类型正确。

于 2013-10-31T10:51:26.043 回答
0

不要尝试编写代码来读取 XML 文档。请改用内置的 XML 序列化程序。http://msdn.microsoft.com/en-us/library/fa420a9y.aspx

于 2013-10-31T10:43:57.300 回答
0 
You can read XML like: 
 XmlDocument doc = new XmlDocument();
 doc.Load(Server.MapPath("\\foldername\\" + "filename.xml"));
 XmlNode node = doc.SelectSingleNode("//Employees//Worker/Salary");
 Response.Write(node.InnerText.ToString());

 This way, you can get 5000 as Salary.
于 2013-10-31T10:47:10.513 回答
0

访问 XML 文件的特定节点的最简单方法是使用 LINQ。

string writePath = Environment.ExpandEnvironmentVariables("%USERPROFILE%") + @"\Desktop";
string writeFile = writePath + @"\Employees.xml";

XDocument xmlDocument = XDocument.Load(writeFile)

那么如果你想读取 Worker 元素中的所有元素

var queryResult = from x in xmlDocument.Root.Element("Worker").Elements() select x;
foreach (var item in queryResult)
{
    Console.WriteLine(item.Value);
}

或者，如果您想单独处理所有 Worker 元素，则只需将查询结果转换为列表

var queryResult = (from x in xmlDocument.Root.Element("Worker").Elements() select x).ToList();
Console.WriteLine("Start reading <ID> element");
Console.WriteLine("Contains: " + queryResult[0]);

Console.WriteLine("Start reading <FirstName> element");
Console.WriteLine("Contains: " + queryResult[1]);

Console.WriteLine("Start reading <LastName> element");
Console.WriteLine("Contains: " + queryResult[2]);

Console.WriteLine("Start reading <Salary> element");
Console.WriteLine("Contains: " + queryResult[3]);
于 2013-10-31T10:56:10.463 回答
0

我发现在对业务对象进行序列化时，使用 XML 总是更容易。

[Serializable]
public class Employees
{
    private List<Worker> _Workers;
    [XmlArray]
    public List<Worker> Workers
    {
        get { return _Workers; }
        set { _Workers = value; }
    }
}

[Serializable]
public class Worker
{
    public Int32 ID { get; set; }
    public String FirstName { get; set; }
    // etc.


    public void SerializeToXML(string outputFolderLocation)
    {
        try
        {
            if (!outputFolderLocation.EndsWith('\\'))
            {
                outputFolderLocation += "\\";
            }

            //Create our own namespaces for the output
            XmlSerializerNamespaces ns = new XmlSerializerNamespaces();

            //Add an empty namespace and empty value
            ns.Add("", "");

            XmlSerializer serializer = new XmlSerializer(typeof(Worker));
            string outpath = outputFolderLocation + "FileName-" + DateTime.Now.ToBinary().ToString() + ".xml";
            XmlTextWriter textWriter = new XmlTextWriter(outpath, Encoding.GetEncoding("ISO-8859-1"));
            serializer.Serialize(textWriter, this, ns);
            textWriter.Close();
        }
        catch (Exception ex)
        {
            throw new Exception("Error serializing to XML", ex);
        }
    }
}

您还可以通过创建值 XML 属性而不是节点来整理您的 XML。为此，您将属性更改为：

[XmlAttribute("ID-Value")]
public Int32 ID { get; set; }
// would serialise like this <worker ID-Value="1"></worker>
于 2013-10-31T10:57:13.347 回答