asp.net 数据网格视图行选择到文本框

asp.net datagrid view row selection into textbox

当用户单击网格视图中的编辑按钮时,我需要帮助将行选择放入文本框,我有事件并将向您展示我尝试过的代码。 我在 visual studio 2013 和 sql 服务器 2012 上 运行。

txtname.Text = gridview.Rows[gridview.SelectedIndex].Cells[1].Text;

txtname.Text = gridview.Rows[gridview.SelectedRow].Cells[1].Text;

[屏幕]http://imgur.com/WCsvnLB,cLdMeQN

而不是 gridview.SelectedIndex 你应该有一个事件参数并获取它的索引。来自 docs:

protected void TaskGridView_RowEditing(object sender, GridViewEditEventArgs e)
  {
    //Set the edit index.
    TaskGridView.EditIndex = e.NewEditIndex;
    //Bind data to the GridView control.
    BindData();
  }

所以你的是:

protected void gridView_RowEditing(object sender, GridViewEditEventArgs e)
{
   txtname.Text = gridview.Rows[e.NewEditIndex].Cells[1].Text;
}

首先使用下面的代码创建一个Gridview

 <asp:GridView ID="GV" runat="server" AutoGenerateColumns="false"  OnRowEditing="GV_RowEditing" DataKeyNames="id">
        <Columns>
            <asp:TemplateField HeaderText="Username">
                <ItemTemplate>
                    <asp:Label ID="Label1" runat="server" Text='<%#Eval("Column1")%>'></asp:Label>
                </ItemTemplate>
            </asp:TemplateField>
            <asp:TemplateField HeaderText="Password">
                <ItemTemplate>
                    <asp:Label ID="Label2" runat="server" Text='<%#Eval("Column2")%>'></asp:Label>
                </ItemTemplate>
            </asp:TemplateField>
            <asp:TemplateField HeaderText="Action">
                <ItemTemplate>
                    <asp:LinkButton ID="LinkButton1" runat="server" CommandName="edit" OnClick="LinkButton1_Click" >Edit</asp:LinkButton>
                </ItemTemplate>
            </asp:TemplateField>
        </Columns>
    </asp:GridView>

然后生成gridview的行编辑事件,在里面写这段代码

 string id = GV.DataKeys[e.NewEditIndex].Value.ToString();
    string select = "select * from tblLogin where id ='"+Convert.ToInt16(id)+"'";
    ds = gs.select(select);
    if (ds.Tables[0].Rows.Count > 0)
    {
       lblName.Text= ds.Tables[0].Rows[0]["Column1"].ToString();
        lblPass.Text=ds.Tables[0].Rows[0]["Column2"].ToString();
    }

希望这会有所帮助

所以我结合了你们的两个答案,想出了一个效果很好的答案:

int id = Convert.ToInt32(gridview.DataKeys[e.NewEditIndex].Value);
    txtid.Text = id.ToString();
    txtname.Text = gridview.Rows[id].Cells[3].Text;
    txtadd.Text = gridview.Rows[id].Cells[4].Text;
    txtcountry.Text = gridview.Rows[id].Cells[5].Text;
    txtcity.Text = gridview.Rows[id].Cells[6].Text;
    txtpin.Text = gridview.Rows[id].Cells[7].Text;

让我知道您的想法,如果这是糟糕的编码也请告诉我。

谢谢