下拉列表选择事件在GridView中出现问题 [英] Problem in gridview at dropdown list select event

问题描述

以下是我编写的代码:-

Following is the code that I have written:-

protected void DropDownList1_SelectedIndexChanged(object sender, EventArgs e)
  {
      SqlCommand com1 = new SqlCommand("quot;select sno,adv,city from adv1 where categry=@b",con);
      com1.Parameters.AddWithValue("@b", DropDownList1.Text);
      con.Open();
      GridView1.DataSource = com1.ExecuteReader();
      con.Close();
      GridView1.DataBind();
}

在网格视图下，我有以下内容:

And under grid view I have the following:

protected void GridView1_PageIndexChanging(object sender, GridViewPageEventArgs e)
    {
        GridView1.PageIndex = e.NewPageIndex;
        GridView1.AllowPaging = true;
    }

它给出以下错误:数据源不支持服务器端数据分页."
有人可以提供帮助吗?

It gives this error: "The data source does not support server-side data paging."
anyone can help?

推荐答案

要进行分页，您需要使用DataAdapter. ExecuteReader 仅是转发的连接数据形式.因此，您将无法进行服务器端分页.

更新:
类似的讨论:链接1 [ ^ ]，链接2 [ ^ ]

试试类似的东西:

In order to have paging you need to use DataAdapter. ExecuteReader is forward only, connected data form. Thus you won''t be able to have server side pagin.

UPDATE:
Similar discussions: Link 1[^], Link 2[^]

Try something like:

DataTable dt = new DataTable();       
SqlDataAdapter ad = new SqlDataAdapter(cmd);
ad.Fill(dt);
GridView1.DataSource = dt;
GridView1.DataBind();

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您可以尝试以下方法:

You can try as bellow:

protected void DropDownList1_SelectedIndexChanged(object sender, EventArgs e)
  {
      SqlCommand com1 = new SqlCommand("select sno,adv,city from adv1 where categry=@b",con);
      com1.Parameters.AddWithValue("@b", DropDownList1.Text);
      con.Open();

      SqlDataAdapter da=new SqlDataAdapter(com1); 
      DataSet ds=new DataSet();
      da.Fill(ds);
      GridView1.DataSource = ds.Tables[0];
      GridView1.DataBind();
      con.Close();
}

protected void GridView1_PageIndexChanging(object sender, GridViewPageEventArgs e)
    {
        //GridView1.PageIndex = e.NewPageIndex; //with or without
        GridView1.AllowPaging = true;
    }

谢谢，
马蒙

Thanks,
Mamun

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