<asp:GridView ID="grdview" runat="server" AutoGenerateColumns="false" OnRowCommand="grdview_RowCommand">
        <Columns>
            <asp:BoundField ReadOnly="True" HeaderText="sno" InsertVisible="False" DataField="sno"

                SortExpression="sno"></asp:BoundField>
            <asp:TemplateField>
                <ItemTemplate>
                    <asp:LinkButton ID="LinkButton2" runat="server" CommandArgument='<%#Eval("site")%>'

                        Text='<%#Eval("site")%>' CommandName='site' CausesValidation="False"></asp:LinkButton>
                </ItemTemplate>
            </asp:TemplateField>
        </Columns>
    </asp:GridView>

protected



ASPX .CS

void Page_Load（object sender，EventArgs e）

{

if（！IsPostBack）

{

con.Open（）;

SqlCommand cmd = new SqlCommand（ select * from newtable，con）;

SqlDataAdapter dtAdapter = new SqlDataAdapter（cmd）;

DataSet ds = new DataSet（）;

dtAdapter.Fill（ds）;

con.Close（）;

grdview.DataSource = ds.Tables [0];

grdview。 DataBind（）;

}

}

protected void grdview_RowCommand（object sender，GridViewCommandEventArgs e）

{

if（e.CommandName ==site）

{

string url =http：//+ e.CommandArgument.ToString （）;

Response.Write（< script> window.open（''+ url +''）;< / script>）;

}



}

protected

ASPX.CS
void Page_Load(object sender, EventArgs e)
{
if (!IsPostBack)
{
con.Open();
SqlCommand cmd = new SqlCommand("select * from newtable", con);
SqlDataAdapter dtAdapter = new SqlDataAdapter(cmd);
DataSet ds = new DataSet();
dtAdapter.Fill(ds);
con.Close();
grdview.DataSource = ds.Tables[0];
grdview.DataBind();
}
}
protected void grdview_RowCommand(object sender, GridViewCommandEventArgs e)
{
if (e.CommandName == "site")
{
string url = "http://" + e.CommandArgument.ToString();
Response.Write("<script>window.open(''" + url + "'' );</script>");
}

}

HI，



绑定Gridview通常就像每个人一样。将链接按钮html设置如下。

bind the Gridview normally, as every body does. Make the link button html as below.

<asp:linkbutton id="YourID" runat="server" autopostback="false" text='<%#Eval("URL")%>' onclientclick="return OpenNewTab(this)" />

并使用如下的javascript函数。

and use the javascript function like below.

function OpenNewTab(sender)
{
    if(sender.innerText !="")
    {
        return window.open(sender.innerText,'_blank');
    }
    else
    {
        return false;
    }
}

希望有帮助

hope it helps

将数据访问层作为类文件实际区分您的表示类和代码逻辑。



在类文件中创建一个方法，如下所示：

Hi,

Have a data access layer as a class file to actually differentiate between your presentation class and your code logics.

Create a method in your class file as below:

public DataTable PopulateGrid()
{
//Have a connection object here, name it as xconn...
SqlDataAdapter da=new SqlDataAdapter("select id as slNo,sitename from dbo.Table",xconn);
DataTable dt=new DataTable();
da.Fill(dt);
return dt;
}

然后在你的aspc.cs文件中，你可能想要在Page_Load（）上填充网格，

then in your aspc.cs file, you might want to populate your grid on Page_Load(),

protected void Page_Load()
{
//create an object of your data access layer
DAL obj=new DAL();
DataTable dt=obj.PopulateGrid();
gridview.DataSource=dt;
gridView.DataBind();
}

使用上述代码，您可以在Page_Load（）上填充网格。

当您单击gridview时重定向到站点，您实际上可以设置所需列的PostBackUrl属性，它将起到作用。



希望它有帮助。



-Anurag

Using the above codes you will be able to populate your grid on Page_Load().
For redirecting to the site when you click your gridview, you can actually set the PostBackUrl property of the required column, and it will do the trick.

Hope it helps.

-Anurag