Question

我已经动态创建了5个下拉列表。它来自aspx.cs

for (int i = 0; i < 5; i++)
            {
                DropDownList drop = new DropDownList();
                drop.ID = "dropdownlist" + i;
                form1.Controls.Add(drop);
                form1.Controls.Add(new LiteralControl("<br />"));
             }

我已经有了另一个下拉代码，比如

if (!this.IsPostBack)
        {
            string constr = ConfigurationManager.ConnectionStrings["constr"].ConnectionString;
            using (SqlConnection con = new SqlConnection(constr))
            {
                using (SqlCommand cmd = new SqlCommand("SELECT ID, Name FROM RejectedProduct"))
                {
                    cmd.CommandType = CommandType.Text;
                    cmd.Connection = con;
                    con.Open();
                    DropDownList1.DataSource = cmd.ExecuteReader();
                    DropDownList1.DataTextField = "Name";
                    DropDownList1.DataValueField = "ID";
                    DropDownList1.DataBind();
                    con.Close();
                }
            }
            DropDownList1.Items.Insert(0, new ListItem("Select Item for adding", "0"));}

如何使用此代码动态创建新的5个下拉列表？

Answer 1

请查看Mysterio11＆amp;的评论。取胜。

填写数据的基本思想如下。最重要的是，它没有得到优化。在循环中创建命令和连接是个坏主意。

        if (!this.IsPostBack)
        {
            DropDownList DropDownList1;
            for (int i = 0; i < 5; i++)
            {
                DropDownList1 = (DropDownList)FindControl("dropdownlist" + i);
                string constr = ConfigurationManager.ConnectionStrings["constr"].ConnectionString;
                using (SqlConnection con = new SqlConnection(constr))
                {
                    using (SqlCommand cmd = new SqlCommand("SELECT ID, Name FROM RejectedProduct"))
                    {
                        cmd.CommandType = CommandType.Text;
                        cmd.Connection = con;
                        con.Open();
                        DropDownList1.DataSource = cmd.ExecuteReader();
                        DropDownList1.DataTextField = "Name";
                        DropDownList1.DataValueField = "ID";
                        DropDownList1.DataBind();
                        con.Close();
                    }
                }
                DropDownList1.Items.Insert(0, new ListItem("Select Item for adding", "0"));
            }
        }

如何在ASP.NET中动态创建下拉控件

1 个答案: