我正在使用下面的代码来绑定另一个表中的下拉数据。并且还使用 rowindex 引用该控件名称。但它总是返回null。并且还返回错误消息。
`Object reference not set to an instance of an object.`
我使用这两种方法,但都返回控件名称 null
第一个代码:
protected void GridView2_RowDataBound(object sender, GridViewRowEventArgs e)
{
if (e.Row.RowType == DataControlRowType.DataRow)
{
Control ctrl = e.Row.FindControl("DDL_STATUS_FT"); //It always return null
if (ctrl != null)
{
DropDownList dd = ctrl as DropDownList;
DataSet7TableAdapters.sp_getall_trv_masterTableAdapter TA = new DataSet7TableAdapters.sp_getall_trv_masterTableAdapter();
DataSet7.sp_getall_trv_masterDataTable DS = TA.GetData();
dd.DataTextField = "fld_TName";
dd.DataValueField = "fld_id";
dd.DataSource = DS;
dd.DataBind();
}
}
}
第二 :
In databind function
if (DS.Rows.Count > 0)
{
GridView2.DataSource = DS;
GridView2.DataBind();
foreach (GridViewRow grdRow in GridView2.Rows)
{
DataSet7TableAdapters.sp_getall_trv_masterTableAdapter TA1 = new DataSet7TableAdapters.sp_getall_trv_masterTableAdapter();
DataSet7.sp_getall_trv_masterDataTable DS1 = TA1.GetData();
// Nested DropDownList Control reference is passed to the DrdList object. This will allow you access the properties of dropdownlist placed inside the GridView Template column.
DropDownList drdList = (DropDownList)(GridView2.Rows[grdRow.RowIndex].Cells[4].FindControl("DDL_STATUS_FT"));//It always return null
// DataBinding of nested DropDownList Control for each row of GridView Control.
drdList.DataSource = DS1;
drdList.DataValueField = "fld_id";
drdList.DataTextField = "fld_TName";
drdList.DataBind();
}
}
请帮我这样做..
<asp:TemplateField ItemStyle-Width="100px" HeaderText="TYPE">
<ItemTemplate>
<asp:DropDownList ID="DDL_STATUS" runat="server" AutoPostBack="true" Enabled="false" >
</asp:DropDownList>
</ItemTemplate>
<EditItemTemplate>
<asp:DropDownList ID="DDL_edit_STATUS" runat="server" AutoPostBack="true" SelectedValue='<%# Eval("fld_Type") %>'>
</asp:DropDownList>
</EditItemTemplate>
<FooterTemplate>
<asp:DropDownList ID="DDL_STATUS_FT" runat="server" AutoPostBack="true">
</asp:DropDownList>
</FooterTemplate>
</asp:TemplateField>