我有一个地方类定义如下:
public class place
{
public string placeID { get; set; }
public string placeCatID { get; set; }
public string placeName { get; set; }
}
数据存储在列表调用地点列表中
List<place> placelist = new List<place>();
可以指导我如何将placelist转换为字符串或将其导出并将其保存为XML文件格式,如下所示:
<place>
<pID>0001</pID>
<pCatID>C1</pID>
<pName>Location 1</pName>
</place>
<place>
<pID>0002</pID>
<pCatID>C1</pID>
<pName>Location 2</pName>
</place>
语言使用是C#
感谢。
答案 0 :(得分:4)
您可以使用任何序列化为XML的序列化程序。我建议使用DataContractSerializer
来自MSDN:
DataContractSerializer s = new DataContractSerializer(typeof(T));
using (FileStream fs = File.Open("test" + typeof(T).Name + ".xml", FileMode.Create))
{
Console.WriteLine("Testing for type: {0}", typeof(T));
s.WriteObject(fs, obj);
}
http://msdn.microsoft.com/en-us/library/bb675198.aspx
在您的情况下,用List<T>
答案 1 :(得分:2)
简而言之,可以使用以下选项:
答案 2 :(得分:2)
下面的解决方案适用于XmlSerializer
,但您也可以使用DataContractSerializer
。
默认情况下,XmlSerializer
会序列化所有字段,而DataContractSerializer
需要明确指定要序列化的内容。
添加序列化属性以具有自定义元素名称:
public class place
{
[XmlElement("pID")]
public string placeID { get; set; }
[XmlElement("pCatID")]
public string placeCatID { get; set; }
[XmlElement("pName")]
public string placeName { get; set; }
}
序列化代码:
var ser = new XmlSerializer(typeof(List<place>));
TextWriter writer = new StreamWriter(@"C:\1.xml");
// o is List<place> here
ser.Serialize(writer, o);
XML:
<?xml version="1.0" encoding="utf-8"?>
<ArrayOfPlace xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<place>
<pID>1</pID>
<pCatID>2</pCatID>
<pName>3</pName>
</place>
<place>
<pID>3</pID>
<pCatID>4</pCatID>
<pName>5</pName>
</place>
</ArrayOfPlace>
答案 3 :(得分:0)
您可以编写自己的ToXml()
方法,也可以使用DataContractSerializer
。
答案 4 :(得分:0)
我刚写了a blog post on saving an object's data to Binary, XML, or Json。因为您希望类变量在xml文件中具有与在代码中不同的名称,所以您需要使用[XmlElement(“NameToShowUpInXmlFileGoesHere”)]来装饰每个公共属性。
完成后,只需调用以下函数即可将对象实例保存到文件或从文件加载。
注意:这需要将System.Xml程序集包含在项目中。
/// <summary>
/// Writes the given object instance to an XML file.
/// <para>Only Public properties and variables will be written to the file. These can be any type though, even other classes.</para>
/// <para>If there are public properties/variables that you do not want written to the file, decorate them with the [XmlIgnore] attribute.</para>
/// <para>Object type must have a parameterless constructor.</para>
/// </summary>
/// <typeparam name="T">The type of object being written to the file.</typeparam>
/// <param name="filePath">The file path to write the object instance to.</param>
/// <param name="objectToWrite">The object instance to write to the file.</param>
/// <param name="append">If false the file will be overwritten if it already exists. If true the contents will be appended to the file.</param>
public static void WriteToXmlFile<T>(string filePath, T objectToWrite, bool append = false) where T : new()
{
TextWriter writer = null;
try
{
var serializer = new XmlSerializer(typeof(T));
writer = new StreamWriter(filePath, append);
serializer.Serialize(writer, objectToWrite);
}
finally
{
if (writer != null)
writer.Close();
}
}
/// <summary>
/// Reads an object instance from an XML file.
/// <para>Object type must have a parameterless constructor.</para>
/// </summary>
/// <typeparam name="T">The type of object to read from the file.</typeparam>
/// <param name="filePath">The file path to read the object instance from.</param>
/// <returns>Returns a new instance of the object read from the XML file.</returns>
public static T ReadFromXmlFile<T>(string filePath) where T : new()
{
TextReader reader = null;
try
{
var serializer = new XmlSerializer(typeof(T));
reader = new StreamReader(filePath);
return (T)serializer.Deserialize(reader);
}
finally
{
if (reader != null)
reader.Close();
}
}
public class place
{
[XmlElement("pID")]
public string placeID { get; set; }
[XmlElement("pCatID")]
public string placeCatID { get; set; }
[XmlElement("pName")]
public string placeName { get; set; }
}
// To write the placeList variable contents to XML.
WriteToXmlFile<List<place>>("C:\places.txt", placeList);
// To read the xml file contents back into a variable.
List<place> placeList= ReadFromXmlFile<List<place>>("C:\places.txt");