SQL Server未显示任何输出

时间:2016-03-15 11:03:26

标签: asp.net sql-server

我正在使用此代码验证用户,但它始终显示“未找到用户”#39;。我是否使用带有where子句的完整select命令。答案每次都一样。虽然它总体上显示了一些结果,但我无法匹配一个结果。

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.UI;
using System.Web.UI.WebControls;
using System.Data;
using System.Data.SqlClient;
using System.Text;
using System.Configuration;

public partial class Applicant : System.Web.UI.Page
{
    protected void Page_Load(object sender, EventArgs e)
    {
        using (SqlConnection cn = new SqlConnection(myConnString))
        {
            using (SqlCommand cmd = new SqlCommand())
            {
                cn.Open();
                SqlDataReader conReader = null;
                cmd.CommandText = "Select * from Applicant ";
                cmd.Connection = cn;
                cmd.CommandType = CommandType.Text;
                // cmd.Parameters.AddWithValue("@userName", myid);
                // cmd.Parameters.AddWithValue("@UserPassword", mypass);
                try
                {
                    conReader = cmd.ExecuteReader();
                    bool _userfound = false;
                    while (conReader.Read())
                    {
                        if (conReader[0].ToString() == myid.ToString() && conReader[1].ToString() == mypass.ToString())
                        {
                            _userfound = true;
                            break;
                        }
                    }
                    if (_userfound)
                        Response.Write("User Found");
                    else
                        Response.Write("User not Found");
                }
                catch (Exception ex)
                {
                    Console.Write(ex);
                }
                finally
                {
                    cn.Close();
                }
            }
        }
    }
}

1 个答案:

答案 0 :(得分:3)

您应该完成查询,例如:

cmd.CommandText = "Select * from Applicant where UserName = @userName";

只有这样才能在查询中添加一个(同名的)参数。