下拉列表显示意外行为

Question

Asp.net 代码：

     Select Employee : </b>
    <asp:DropDownList ID="DropDownListSelectEmployee" runat="server" AutoPostBack="true" 
OnSelectedIndexChanged="DropDownListSelectEmployee_SelectedIndexChanged"
 OnTextChanged="DropDownListSelectEmployee_TextChanged"  >
                                    </asp:DropDownList>

后面的代码：

  public void DropDownListSelectEmployee_Fill()
        {
            DropDownListSelectEmployee.Items.Clear();
            string q = "select username from nworksuser where _type='Employee' or _type='Admin_Employee';";
            MySqlCommand cmd = new MySqlCommand(q, conn);
            conn.Open();
            string user = "";
            MySqlDataReader rdr = cmd.ExecuteReader();
            while (rdr.Read())
            {
                user = rdr.GetString("username");
                DropDownListSelectEmployee.Items.Add(user);
            }
            conn.Close();
        }
 protected void DropDownListSelectEmployee_TextChanged(object sender, EventArgs e)
        {
            string q = "select uid,fname,mname,lname,date_format(hire_date,'%Y-%m-%d'),desg,date_format(dob,'%Y-%m-%d'),AddressLine1,AddressLine2,state,city,pincode,country,MobileNo,UserActive,username,date_format(annivarsary,'%Y-%m-%d'),email from nWorksUser where Username='" + DropDownListSelectEmployee.Text + "'";
            MySqlCommand cmd = new MySqlCommand(q, conn);
            conn.Open();
            MySqlDataAdapter dataAdapter = new MySqlDataAdapter(q, conn);
            DataTable dt = new DataTable();
            dataAdapter.Fill(dt);
            foreach (DataRow dr in dt.Rows)
            {

                txtFirstName.Text = dr[1].ToString();
                txtMiddleName.Text = dr[2].ToString();
                txtLastName.Text = dr[3].ToString();
                txtMobile.Text = dr[13].ToString();
                txtUsername.Text = dr[15].ToString();
                txtState.Text = dr[9].ToString();
                txtPincode.Text = dr[11].ToString();
                txtEmail.Text = dr[17].ToString();
                txtDob.Text = dr[6].ToString();
                txtDesignation.Text = dr[5].ToString();
                txtDateofHire.Text = dr[4].ToString();
                txtDateofAnnivarsary.Text = dr[16].ToString();
                txtCountry.Text = dr[12].ToString();
                txtCity.Text = dr[10].ToString();
                txtAddressLine2.Text = dr[7].ToString();
                txtAddressLine1.Text = dr[8].ToString();

                if (dr[14].ToString().Length == 4)
                {
                    RdoGender.SelectedIndex = 1;
                }
                else
                {
                    RdoGender.SelectedIndex = 2;
                }
                conn.Close();
            }

        }
 protected void DropDownListSelectEmployee_SelectedIndexChanged(object sender, EventArgs e)
        {

        }

下拉列表包含数据库中的用户名。现在我想要这样，当您从下拉列表中选择任何用户时，它应该以表单及其工作方式显示他的详细信息。但问题是，在选择任何一个用户后，该名称会暂时出现在下拉列表中，并将其重置为下拉列表的第一个名称，任何显示其相关而不是选择的详细信息。它不接受除名字下拉列表之外的任何其他名称。我不知道问题出在哪里。请帮帮我。

Answer 1

您需要用if (!Page.IsPostBack) 将DropDownListSelectEmployee_Fill 中的代码包装起来。在回帖时，下拉内容将重新填充，导致其状态重置，这就是您看到“意外行为”的原因。

检查回发，您可以确定下拉菜单不会受到选择更改的影响：

public void DropDownListSelectEmployee_Fill()
{
    if (!Page.IsPostBack) 
    {
        DropDownListSelectEmployee.Items.Clear();
        string q = "select username from nworksuser where _type='Employee' or _type='Admin_Employee';";
        MySqlCommand cmd = new MySqlCommand(q, conn);
        conn.Open();
        string user = "";
        MySqlDataReader rdr = cmd.ExecuteReader();
        while (rdr.Read())
        {
            user = rdr.GetString("username");
            DropDownListSelectEmployee.Items.Add(user);
        }
        conn.Close();
    }
}