C# 显示登录用户的详细信息

我目前正在开发一个web应用程序，我想显示当前登录用户的详细信息。当用户登录时，我已经创建了一个新会话，并尝试使用它来验证当前用户并显示其详细信息。C代码如下所示

protected void Page_Load(object sender, EventArgs e)
    {
        try
        {
            SqlConnection conn = new SqlConnection(ConfigurationManager.ConnectionStrings["ConnectionString"].ConnectionString);
            conn.Open();
            string detailsQuery = "select * FROM [Customer] where Customer_No ='" + Session["New"] + "'";
            SqlCommand com = new SqlCommand(detailsQuery, conn);
            com.ExecuteNonQuery();                
            Response.Write("Details Showing");
            conn.Close();
        }
        catch (Exception ex)
        {
            Response.Write("Error:" + ex.ToString());
        }
    }

---

我可以很容易地显示所有客户的详细信息，但我不知道如何显示当前登录的用户。如果有任何帮助，我将不胜感激。

您需要将数据输入UI。控制使这变得容易。它可以根据绑定到它的数据自动生成HTML表。我们将在标记中声明一个GridView，将数据选择为一个，然后将DataTable绑定到GridView

<asp:GridView runat="server" id="CustomerDetailsGV" AutoGenerateColumns="true" />

您还应该使用

语句将
SqlConnection
包装在
中

string detailsQuery = "select * FROM [Customer] where Customer_No ='" + Session["New"] + "'";
SqlCommand com = new SqlCommand(detailsQuery);
DataTable dt = new DataTable();
using (var conn = new SqlConnection(ConfigurationManager.ConnectionStrings["ConnectionString"].ConnectionString)
{
    com.Connection = conn;
    conn.Open();                
    dt.Load(com.ExecuteReader());                       
}
CustomerDetailsGV.DataSource = dt;
CustomerDetailsGV.DataBind();


您可能会发现这是一个有用的功能：

/// <summary>
/// Executes a database command with the specified connection and returns a data table synchronously.
/// </summary>
/// <param name="command">The command to execute.</param>
/// <param name="connection">The connection to use.</param>
/// <returns>A DataTable representing the command results.</returns>
public static DataTable GetDataTable(SqlCommand command, SqlConnection connection)
{
    DataTable dt = new DataTable();
    command.Connection = connection;
    using (connection)
    {
        connection.Open();
        dt.Load(command.ExecuteReader());
    }
    return dt;
}

您网页上的每个人都可以通过Session.SessionID字符串进行标识。
.  在Customer表中，添加SessionID字段（SQL Varchar（70））。当客户登录时，更新该客户的session.sessionID。如果要在屏幕上显示当前客户的信息，只需提供他们的Session.sessionID即可

protected void Page_Load(object sender, EventArgs e)
{
    try
    {
        SqlConnection conn = new SqlConnection(ConfigurationManager.ConnectionStrings["ConnectionString"].ConnectionString);
        conn.Open();
        string detailsQuery = "select * FROM [Customer] where SessionID ='" + Session.SessionID + "'";
        SqlCommand com = new SqlCommand(detailsQuery, conn);
        com.ExecuteNonQuery();                
        Response.Write("Details Showing");
        conn.Close();
    }
    catch (Exception ex)
    {
        Response.Write("Error:" + ex.ToString());
    }
}

我不明白你的问题是什么。您说“我可以轻松地显示所有客户的详细信息”，但您没有任何代码来执行此操作。我可以使用工具箱visual studio中的表格向导来执行此操作。非常感谢您的帮助。@colliec啊，最后一件事。加载数据时，您应该在
if（！IsPostBack）{/*yourcodehere*/}
块中完成所有操作。否则，回发时，数据将被重新加载，这可能不是您想要的。@AndrewGrinder您应该使用
@
语法将该注释定向到colliec。您还应该使用可在此处查看的参数化查询：dotnetperls.com/sqlparameter。这将避免任何SQL注入攻击。“代码就像每个人都是小偷。”-Bawan在codelesscode.com/case/140（Heartbleed）中再次谈到参数化查询：一个查询问题的旧答案：@colliec
string detailsQuery = "select * FROM [Customer] where Customer_No ='" + Session["New"] + "'";
SqlCommand com = new SqlCommand(detailsQuery);
SqlConnection conn = new SqlConnection(ConfigurationManager.ConnectionStrings["ConnectionString"].ConnectionString);
CustomerDetailsGV.DataSource = GetDataTable(com, con);
CustomerDetailsGV.DataBind();

protected void Page_Load(object sender, EventArgs e)
{
    try
    {
        SqlConnection conn = new SqlConnection(ConfigurationManager.ConnectionStrings["ConnectionString"].ConnectionString);
        conn.Open();
        string detailsQuery = "select * FROM [Customer] where SessionID ='" + Session.SessionID + "'";
        SqlCommand com = new SqlCommand(detailsQuery, conn);
        com.ExecuteNonQuery();                
        Response.Write("Details Showing");
        conn.Close();
    }
    catch (Exception ex)
    {
        Response.Write("Error:" + ex.ToString());
    }
}