我有List<>
我已设法写入文件。
现在我正在尝试读取同一个文件并将其写回List<>
。
有没有办法做到这一点?
任何人都可以帮我开始吗?
会有用吗?
答案 0 :(得分:9)
我认为最简单的方法是使用XmlSerializer
:
XmlSerializer serializer = new XmlSerializer(typeof(List<MyClass>));
using(FileStream stream = File.OpenWrite("filename"))
{
List<MyClass> list = new List<MyClass>();
serializer.Serialize(stream, list);
}
using(FileStream stream = File.OpenRead("filename"))
{
List<MyClass> dezerializedList = (List<MyClass>)serializer.Deserialize(stream);
}
答案 1 :(得分:7)
您可以尝试此操作(使用System.Xml.Linq)
XDocument xmlDoc = XDocument.Load("yourXMLFile.xml");
var list = xmlDoc.Root.Elements("id")
.Select(element => element.Value)
.ToList();
答案 2 :(得分:5)
您可以使用LINQ to XML读取XML文件并将其绑定到List。
http://www.mssqltips.com/sqlservertip/1524/reading-xml-documents-using-linq-to-xml/此链接有足够的信息。
这是我过去所做的事情;我希望它有所帮助。我想你想要完全相同的事情 public static List<ProjectMap> MapInfo()
{
var maps = from c in XElement.Load(System.Web.Hosting.HostingEnvironment.MapPath("/ProjectMap.xml")).Elements("ProjectMap")
select c;
List<ProjectMap> mapList = new List<ProjectMap>();
foreach (var item in maps)
{
mapList.Add(new ProjectMap() { Project = item.Element("Project").Value, SubProject = item.Element("SubProject").Value, Prefix = item.Element("Prefix").Value, TableID = item.Element("TableID").Value });
}
return mapList;
}
答案 3 :(得分:4)
一种简单的方法是
using System;
using System.Linq;
using System.Xml.Linq;
public class Test
{
static void Main()
{
string xml = "<Ids><id>1</id><id>2</id></Ids>";
XDocument doc = XDocument.Parse(xml);
List<string> list = doc.Root.Elements("id")
.Select(element => element.Value)
.ToList();
}
}
答案 4 :(得分:0)
如果您正在使用Singleton模式,那么就是如何将XML读入其中!
public static GenericList Instance {
get {
XElement xelement = XElement.Load(HostingEnvironment.MapPath("RelativeFilepath"));
IEnumerable<XElement> items = xelement.Elements();
instance = new GenericList();
instance.genericList = new List<GenericItem>{ };
foreach (var item in items) {
//Get the value of XML fields here
int _id = int.Parse(item.Element("id").Value);
string _name = item.Element("name").Value;
instance.genericList.Add(
new GenericItem() {
//Load data into your object
id = _id,
name = _name
});
}
return instance;
}
}
这会打开CRUD辅助功能,在写入xml
时更新有点棘手 public void Save() {
XDocument xDoc = new XDocument(new XDeclaration("Version", "Unicode type", null));
XElement root = new XElement("GenericList");
//For this example we are using a Schema to validate our XML
XmlSchemaSet schemas = new XmlSchemaSet();
schemas.Add("", HostingEnvironment.MapPath("RelativeFilepath"));
foreach (GenericItem item in genericList) {
root.Add(
//Assuming XML has a structure as such
//<GenericItem>
// <name></name>
// <id></id>
//</GenericItem>
new XElement("GenericItem",
new XElement("name", item.name),
new XElement("id", item.id)
));
}
xDoc.Add(root);
//This is where the mentioned schema validation takes place
string errors = "";
xDoc.Validate(schemas, (obj, err) => {
errors += err.Message + "/n";
});
StringWriter writer = new StringWriter();
XmlWriter xWrite = XmlWriter.Create(writer);
xDoc.Save(xWrite);
xWrite.Close();
if (errors == "")
{
xDoc.Save(HostingEnvironment.MapPath("RelativeFilepath"));
}
}