asp.net read XML code (1/2)

Provides the simplest ASP tutorial. NET Read XML code program

New XDocument (
New XElement ("Ol",
From item in Xdocument.load (Http://111cn.net/update.xml). Descendants ("item")
Select New xelement[]{
New XElement ("Li",
New XElement ("A", (string) item.element ("title"),
New XAttribute ("href", (string) item.element ("link")
)
)
}
)
). WriteTo (New XmlTextWriter (Response.outputstream,encoding.utf8));

Method One: Using an XML control
<% @ Page language= "C #"%>

<body>
<from runat=server>
<asp Tutorial: xml id= "XML1" documentsource= "Grade.xml" runat= "Server"/>
</from>

</body>

Method two: Using DOM technology
<% @ Page language= "C #"%>
<% @ import namespace= "System.Xml"%>
<% @ import namespace= "System.Xml.Xsl"%>

<script language= "C #" runat= "Server" >
void Page_Load (Object Sender,eventargs e)
{
XmlDocument doc=new XmlDocument ();
Doc.load (Server.MapPath ("Grade.xml"));
Xml1.document=doc;
}
</script>
<body>
<from runat=server>
<asp:xml id= "XML1" runat= "Server"/>
</from>

</body>

Method Three: Working with a DataSet object
<% @ Page language= "C #"%>
<% @ import namespace= "System.Data"%>
<% @ import namespace= "System.Data.OleDb"%>
<script language= "C #" runat= "Server" >
void Page_Load (Object Sender,eventargs e)
{
DataSet objdataset=new DataSet ();
Objdataset.readxml (Server.MapPath ("Grade.xml"));
dgemployees.datasource=objdataset.tables["Student"].defaultview;
Dgemployees.databind ();
}
</script>
<body>
<asp:datagrid id= "dgemployees" runat= "Server"/>

</body>

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