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我有以下代碼:使動態創建下拉列表公開的其他類
foreach (ListItem item in check.Items)
{
if (item.Selected)
{
TableRow row = new TableRow();
row.Style.Add("color", "white");
TableCell cell = new TableCell();
cell.Style.Add("background-color", "blue");
Label lbl = new Label();
lbl.Text = item.Text;
cell.Controls.Add(lbl);
row.Cells.Add(cell);
TableCell cell1 = new TableCell();
cell1.Style.Add("background-color", "green");
DropDownList drop = new DropDownList();
drop.Style.Add("align", "right");
drop.Items.Add(" ");
drop.Items.Add("1");
drop.Items.Add("2");
drop.DataValueField = "0";
drop.DataValueField = "1";
drop.DataValueField = "2";
cell1.Controls.Add(drop);
row.Cells.Add(cell1);
this.TblCheck.Rows.Add(row);
drop.SelectedIndexChanged += new EventHandler(drop_SelectedIndexChanged);
drop.AutoPostBack = true;
}
}
}
private void drop_SelectedIndexChanged(object sender,EventArgs e)
{
if (drop.SelectedValue == "1")
{
TableRow row = new TableRow();
TableCell cell = new TableCell();
cell.Style.Add("background-color", "blue");
Label lbl = new Label();
lbl.Text = "H1";
cell.Controls.Add(lbl);
row.Cells.Add(cell);
this.tabel1.Rows.Add(row);
}
if (drop.SelectedValue == "2")
{
TableRow row = new TableRow();
TableCell cell = new TableCell();
cell.Style.Add("background-color", "blue");
Label lbl = new Label();
lbl.Text = "H1";
cell.Controls.Add(lbl);
row.Cells.Add(cell);
TableRow row1 = new TableRow();
TableCell cell1 = new TableCell();
cell1.Style.Add("background-color", "blue");
Label lbl1 = new Label();
lbl1.Text = "H2";
cell1.Controls.Add(lbl1);
row1.Cells.Add(cell1);
this.tabel1.Rows.Add(row);
this.tabel1.Rows.Add(row1);
}
}
我有一個下拉列表,當我爲每個選項選擇的選項中選擇一個行創建了包含選定值,另一列有一個dropdownlist.I要爲該下拉列表創建動態添加一個selectedindexchanged.I請求如何在我selectedindexchange中將引用更改爲動態創建的下拉列表,並將其命名爲「drop」。我試圖讓它公開,但不是我的表格不是很好創建。它是一個使用c#的as.net web應用程序。
sender是動態創建的下拉列表的名稱? – Bibu 2012-03-30 14:10:45
'sender'是觸發事件的對象。在你的情況下,它將是你關聯到'drop_SelectedIndexChanged'方法的動態創建的DropDownList。即你的線'drop.SelectedIndexChanged + =新的EventHandler(drop_SelectedIndexChanged);' – CAbbott 2012-03-30 14:12:37
所以我dropdownlist的名稱下降...如果我寫var drop1 =(DropDownList)下降 - > drop不被識別 – Bibu 2012-03-30 14:16:48