2
这里是问题: 例如,如果我输入“A”作为用户名和“A”作为密码,则单击添加用户没有任何反应。但是如果我再回来,现在输入“B”作为用户名,输入“B”作为密码,然后点击添加用户,它将前一个条目“A”&“A”添加到表格中。喜欢它后面的一个?输入数据(用户名和密码)
我知道我有一些不合适的地方。
public partial class frmManageUsers : System.Web.UI.Page
{
protected void Page_Load(object sender, EventArgs e)
{
}
protected void btnAddUser_Click1(object sender, EventArgs e)
{
//string userName, userPassword;
if (txtUserName.Text == "" || txtUserName.Text == null)
{
lblError.Text = ("User Name may not be empty");
lblError.ForeColor = System.Drawing.Color.Red;
return;
}
// else
// userName = (txtUserName.Text);
if (txtPassword.Text == "" || txtPassword.Text == null)
{
lblError.Text = ("Password may not be empty");
lblError.ForeColor = System.Drawing.Color.Red;
return;
}
//else
// userPassword = (txtPassword.Text);
//using (OleDbConnection conn = new OleDbConnection("PayrollSystem_DBConnectionString"))
OleDbConnection conn =
new OleDbConnection(ConfigurationManager.ConnectionStrings["PayrollSystem_DBConnectionString"].ConnectionString);
{
string insert = "Insert INTO tblUserLogin (UserName, UserPassword, SecurityLevel) Values (@UserName, @UserPassword, @SecurityLevel)";
OleDbCommand cmd = new OleDbCommand(insert, conn);
cmd.Parameters.Add("@UserName", txtUserName.Text);
cmd.Parameters.Add("@UserPassword", txtPassword.Text);
cmd.Parameters.Add("@SecurityLevel", drpdwnlstSecurityLevel.SelectedValue);
conn.Open();
cmd.ExecuteNonQuery();
}
Session["UserName"] = txtUserName.Text;
Session["Password"] = txtPassword.Text;
Session["SecurityLevel"] = drpdwnlstSecurityLevel.SelectedValue;
Server.Transfer("frmManageUsers.aspx");
//Server.Transfer("grdUserLogin");
}
protected void drpdwnlstSecurityLevel_SelectedIndexChanged(object sender, EventArgs e)
{
}
}
你是否第一次使用Server.Transfer进入页面? –