Datagrid class (Generic) ??????

Harish77 · Jan 28, 2004

Hi All,

I am working with a project using ASP.NET. It is an intranet project, and most of the data is displayed using datagrid.

Most of the screens I have Add records, edit records, delete, paging, filtering the data etc.

Now it becomes a repitative task to code for each page the datagrid functionality.

I was wondering if any one of you have come across a datagrid class which can be called whereever necessary and display the result. I checked in the web but I couldnt find anything like that. Please help if you have come across.

Thanks in Advance

Harish

shaddow · Jan 28, 2004

You could write an ascx file that could make what your DataGrid does and then use that control in any page you want with the proper settings.

________
George, M

Harish77 · Jan 28, 2004

Thank you Shaddow, Sorry for asking, but do you have any sample code for doing this.

shaddow · Jan 29, 2004

I've made a simple example

Code:

-LoginCtrl.ascx

<%@ Control Language=&quot;vb&quot; AutoEventWireup=&quot;false&quot; Codebehind=&quot;LoginCtrl.ascx.vb&quot; Inherits=&quot;CustomControl.LoginCtrl&quot; TargetSchema=&quot;[URL unfurl="true"]http://schemas.microsoft.com/intellisense/ie5&quot;[/URL] %>
<asp:Table HorizontalAlign=&quot;Center&quot; id=&quot;loginTable&quot; runat=&quot;server&quot;>
	<asp:TableRow>
		<asp:TableCell HorizontalAlign=&quot;Right&quot;>
			<font face=&quot;verdana&quot; size=&quot;1&quot;>Username:</font>
		</asp:TableCell>
		<asp:TableCell>
			<asp:TextBox ID=&quot;_uname&quot; Runat=&quot;server&quot;></asp:TextBox>
		</asp:TableCell>
	</asp:TableRow>
	<asp:TableRow>
		<asp:TableCell HorizontalAlign=&quot;Right&quot;>
			<font face=&quot;verdana&quot; size=&quot;1&quot;>Password:</font>
		</asp:TableCell>
		<asp:TableCell>
			<asp:TextBox ID=&quot;_password&quot; Runat=&quot;server&quot;></asp:TextBox>
		</asp:TableCell>
	</asp:TableRow>
	<asp:TableRow>
		<asp:TableCell ColumnSpan=&quot;2&quot; HorizontalAlign=&quot;Center&quot;>
			<asp:Button Runat=&quot;server&quot; Text=&quot;Login&quot; ID=&quot;LoginBtn&quot;></asp:Button>
		</asp:TableCell>
	</asp:TableRow>
</asp:Table>

-LoginCtrl.ascx.vb

Public MustInherit Class LoginCtrl
    Inherits System.Web.UI.UserControl
    Public _uname As TextBox
    Public _password As TextBox
#Region &quot; Web Form Designer Generated Code &quot;

    'This call is required by the Web Form Designer.
    <System.Diagnostics.DebuggerStepThrough()> Private Sub InitializeComponent()

    End Sub

    Private Sub Page_Init(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Init
        'CODEGEN: This method call is required by the Web Form Designer
        'Do not modify it using the code editor.
        InitializeComponent()
    End Sub

#End Region
End Class


-WebForm1.aspx

<%@ Page Language=&quot;vb&quot; AutoEventWireup=&quot;false&quot; Codebehind=&quot;WebForm1.aspx.vb&quot; Inherits=&quot;CustomControl.WebForm1&quot;%>
<%@ Register TagPrefix=&quot;uc1&quot; TagName=&quot;LoginCtrl&quot; Src=&quot;LoginCtrl.ascx&quot; %>
<HTML>
	<HEAD>
	</HEAD>
	<body>
		<form id=&quot;Form1&quot; method=&quot;post&quot; runat=&quot;server&quot;>
			<uc1:loginctrl id=&quot;login&quot; runat=&quot;server&quot;></uc1:loginctrl></form>
	</body>
</HTML>

-WebForm1.aspx.vb

Imports CustomControl.LoginCtrl
Public Class WebForm1
    Inherits System.Web.UI.Page
    Protected login As LoginCtrl
#Region &quot; Web Form Designer Generated Code &quot;

    'This call is required by the Web Form Designer.
    <System.Diagnostics.DebuggerStepThrough()> Private Sub InitializeComponent()

    End Sub

    Private Sub Page_Init(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Init
        'CODEGEN: This method call is required by the Web Form Designer
        'Do not modify it using the code editor.
        InitializeComponent()
    End Sub

#End Region

    Private Sub Page_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load
        'Put user code to initialize the page here
        If IsPostBack() Then
            Dim Uname, Pwd As String
            Uname = login._uname.Text
            Pwd = login._password.Text
        End If
    End Sub

End Class

This should be the easyest way of making an Custom Control in my opinion, coding it's clean and simple.

________
George, M

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Datagrid class (Generic) ??????

Harish77

Programmer

shaddow

Programmer

Harish77

Programmer

shaddow

Programmer

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