Asp.net 使用LINQ连接两个数据表,并在其他数据表中获得结果
我有两个数据表,我将它们连接起来以选择一些匹配的记录,我需要从这两个数据表中选择一些列作为一个新的结果集,我就是这样做的Asp.net 使用LINQ连接两个数据表,并在其他数据表中获得结果,asp.net,.net,vb.net,linq,datatable,Asp.net,.net,Vb.net,Linq,Datatable,我有两个数据表,我将它们连接起来以选择一些匹配的记录,我需要从这两个数据表中选择一些列作为一个新的结果集,我就是这样做的 Dim query4 = From x In dtblPolicyFormStopCodes.AsEnumerable() Join y In dtblPolicyFormLetterReq.AsEnumerable() On x.Field(Of String)("Code") Equals y.Field(Of String)("Code"
Dim query4 = From x In dtblPolicyFormStopCodes.AsEnumerable() Join y In dtblPolicyFormLetterReq.AsEnumerable()
On x.Field(Of String)("Code") Equals y.Field(Of String)("Code") _
Select New With _
{ _
.Code = x.Field(Of String)("Code"), _
.Sequence = x.Field(Of Integer)("Sequence "), _
.FieldDataType= x.Field(Of String)("FieldDataType"), _
.FieldValue= y.Field(Of String)("FieldValue") _
}
MyNewDataTable = GetDynamicTableSchema()
query4.Cast(Of DataRow).CopyToDataTable(MyNewDataTable, LoadOption.OverwriteChanges)
但我无法使用两个数据表的选定列获取新数据表,请帮助我如何在“MyNewDataTable”中获取新结果集或作为新数据表获取新结果集
Public Function GetDynamicTableSchema() As System.Data.DataTable
Dim tblDynamicTableDataMain As System.Data.DataTable = Nothing
Try
tblDynamicTableDataMain = New System.Data.DataTable("DynamicTableData")
With tblDynamicTableDataMain
.Columns.Add("Code")
.Columns.Add("Sequence", System.Type.GetType("System.Int32"))
.Columns.Add("FieldDataType")
.Columns.Add("FieldValue")
End With
Return tblDynamicTableDataMain
End Try
End Function
通过使用这种方式,我实现了我想要的功能
Dim dr As DataRow
Dim MyNewDataTable as Datatable = GetDynamicTableSchema()
Dim query = (From x In dtblPolicyFormStopCodes.AsEnumerable() Join y In dtblPolicyFormLetterReq.AsEnumerable()
On x.Field(Of String)("Code") Equals y.Field(Of String)("Code")) _
.Select(Function(a) As DataRow
dr = MyNewDataTable.NewRow()
dr("Code") = a.x.Field(Of String)("Code")
dr("Sequence") = a.x.Field(Of Integer)("Sequence")
dr("FieldDataType") = a.x.Field(Of String)("FieldDataType")
dr("FieldValue") = a.y.Field(Of String)("Fieldvalue")
Return dr
End Function)
MyNewDataTable = query.CopyToDataTable()