如何在.net中解决此错误？

Question

错误：-System.Data.SqlClient.SqlException（0x80131904）：无效的列名称'Username'。

protected void Button1_Click(object sender, EventArgs e)
{
    try
    {

        SqlConnection conn = new SqlConnection(ConfigurationManager.ConnectionStrings["RegistrationConnectionString"].ConnectionString);
        conn.Open();
        string insertQuery = "insert into Registration (Name,Username,Password,Email,Phone) values ('" + tname.Text + "','" + tusename.Text + "','" + tpassword.Text + "','" + temail.Text + "','" + tphonenumber.Text + "')";
        SqlCommand com = new SqlCommand(insertQuery, conn);

        com.Parameters.AddWithValue("@Name",tname.Text);
        com.Parameters.AddWithValue("@Usename",tusename.Text);
        com.Parameters.AddWithValue("@Password",tpassword.Text);
        com.Parameters.AddWithValue("@Email",temail.Text);
        com.Parameters.AddWithValue("@Phone",tphonenumber.Text);


        com.ExecuteNonQuery();
        Response.Redirect("RegistrationDatasave.aspx");
        Response.Write("Registration is successful");
        conn.Close();

    }
    catch (Exception ex)
    {
        Response.Write(ex);
    }
}

Answer 1

如果查询中没有参数，则参数不起作用。

您的查询应该是这样的

using(SqlConnection conn = new SqlConnection(ConfigurationManager.ConnectionStrings["RegistrationConnectionString"].ConnectionString))
{
    conn.Open();
    string insertQuery = "insert into Registration (Name,Username,Password,Email,Phone) values (@Name,@Usename,@Password,@Email,@Phone)";
    SqlCommand com = new SqlCommand(insertQuery, conn);

    com.Parameters.AddWithValue("@Name",tname.Text);
    com.Parameters.AddWithValue("@Usename",tusename.Text);
    com.Parameters.AddWithValue("@Password",tpassword.Text);
    com.Parameters.AddWithValue("@Email",temail.Text);
    com.Parameters.AddWithValue("@Phone",tphonenumber.Text);


    com.ExecuteNonQuery();
    Response.Redirect("RegistrationDatasave.aspx");
    Response.Write("Registration is successful");
}

此外，您需要检查您的注册表，它实际上有一个名为Username的列

Answer 2

我假设您的列名错误。

com.Parameters.AddWithValue("@Usename",tusename.Text);

需要是

com.Parameters.AddWithValue("@Username",tusename.Text);

所以增加了r