ASP.NET C＃從數據庫檢索多個值並添加會話

Question

如何從SQL Server數據庫中添加多個列值並插入session[] ？

只是FirstName正在工作，當我添加LastName ，出現錯誤。

string constr = sql connection here.... ;

using (SqlConnection con = new SqlConnection(constr))
{
    using (SqlCommand cmd = new SqlCommand("SELECT Email, [FirstName] FROM Personal WHERE Email = @email"), cmd2 = new SqlCommand("SELECT Email, [LastName] FROM Personal WHERE Email = @email"))
    {
        cmd.Parameters.AddWithValue("@email", TextboxUsr.Text);
        cmd.Connection = con;
        cmd2.Parameters.AddWithValue("@email", TextboxUsr.Text);
        cmd2.Connection = con;

        con.Open();

        using (SqlDataReader sdr = cmd.ExecuteReader())
        {
            if (sdr.Read())
            {
                string fName = sdr["FirstName"].ToString();
                //string lName = sdr["LastName"].ToString();

                string userFullName = fName;
                Session["New"] = userFullName;
            }
        }

        //con.Close();
    }
}

Answer 1

步驟01：了解有關SQL和C＃基礎知識的更多信息

您需要閱讀更多內容

• SQL SELECT語句

• .net-什么是C＃Using塊，為什么要使用它？ - 堆棧溢出

• c＃-使用語句中的SqlConnection.Close（）-代碼日志

• var vs具體類型用法

• $-字符串插值-C＃參考| 微軟文檔

步驟02：以正確的方式使用“選擇”

• 更改

   SqlCommand cmd = new SqlCommand("SELECT Email, [FirstName] FROM Personal WHERE Email = @email"), cmd2 = new SqlCommand("SELECT Email, [LastName] FROM Personal WHERE Email = @email")) 

• 至

   SqlCommand cmd = new SqlCommand("SELECT Email, [FirstName],[LastName] FROM Personal WHERE Email = @email") 

步驟03：閱讀完整代碼

var sql = @"SELECT Email, [FirstName],[LastName] FROM Personal WHERE Email = @email";
using (var con = new SqlConnection(constr))
using (var cmd = new SqlCommand(sql, con))
{
    con.Open();
    cmd.Parameters.AddWithValue("@email", TextboxUsr.Text);
    using (var sdr = cmd.ExecuteReader())
    {
        if (sdr.Read())
        {
            var fName = sdr["FirstName"].ToString();
            var lName = sdr["LastName"].ToString();
            var userFullName = fName + " " + lName;
            Session["New"] = userFullName;
        }
    }
}