访问基于 XML 的数据
DataSet 支持将 FileStream 对象用作参数的 ReadXml 方法。在这种情况下读取的文件必须同时包含希望读取的架构和数据。DataSet 期望数据在窗体中,如下面的示例所示。
Sub Page_Load(Sender As Object, E As EventArgs)
Dim DS As New DataSet
Dim FS As FileStream
Dim Reader As StreamReader
FS = New FileStream(Server.MapPath("schemadata.xml"),FileMode.Open,FileAccess.Read)
Reader = New StreamReader(FS)
DS.ReadXml(Reader)
FS.Close()
Dim Source As DataView
Source = new DataView(ds.Tables(0))
MyLiteral.Text = Source.Table.TableName
MyDataGrid.DataSource = Source
MyDataGrid.DataBind()
End Sub
<DocumentElement>
<TableName>
<ColumnName1>column value</ColumnName1>
<ColumnName2>column value</ColumnName2>
<ColumnName3>column value</ColumnName3>
<ColumnName4>column value</ColumnName4>
</TableName>
<TableName>
<ColumnName1>column value</ColumnName1>
<ColumnName2>column value</ColumnName2>
<ColumnName3>column value</ColumnName3>
<ColumnName4>column value</ColumnName4>
</TableName>
</DocumentElement>
每个 TableName 节与表中的一行相对应。
Sub Page_Load(Sender As Object, E As EventArgs)
Dim DS As New DataSet
Dim FS As FileStream
Dim Reader As StreamReader
FS = New FileStream(Server.MapPath("schemadata.xml"),FileMode.Open,FileAccess.Read)
Reader = New StreamReader(FS)
DS.ReadXml(Reader)
FS.Close()
Dim Source As DataView
Source = new DataView(ds.Tables(0))
MyLiteral.Text = Source.Table.TableName
MyDataGrid.DataSource = Source
MyDataGrid.DataBind()
End Sub

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