asp.net插入oracle问题

Question

web.config
<connectionStrings>
<add name="conn" connectionString="Data Source=(DESCRIPTION=(ADDRESS=(PROTOCOL=TCP)(HOST=192.168.1.34)(PORT=1521)) (CONNECT_DATA=(SERVICE_NAME=account)));User ID=scott;Password=tiger;Unicode=True" providerName="System.Data.OracleClient"/>
</connectionStrings>
源码OracleConnection oraconn = null;        try
        {
            OracleConnection conn = new OracleConnection(ConfigurationManager.ConnectionStrings["conn"].ConnectionString);            string sql = "insert into amt_account(accountid,password) values(:accountid,:password)";
            OracleCommand cmd = new OracleCommand(sql, oraconn);
            cmd.Parameters.Add("accountid", OracleType.VarChar, 32).Value = txtName;
            cmd.Parameters.Add("password", OracleType.VarChar, 8).Value = txtPass;            cmd.Connection.Open();
            cmd.ExecuteNonQuery();
            Label3.Text = "注册成功";
            cmd.Connection.Close();
        }
        catch (Exception)
        {
            Label3.Text = "注册失败";
            Label3.Style["color"] = "red";
        }
运行以后显示 注册失败 求教各位高手！！！

Answer 1

是不是这样的?  
cmd.Parameters.Add("accountid", OracleType.VarChar, 32).Value = txtName.Text;

Answer 2

 string sql = "insert into amt_account(accountid,password) values"+
 + "(+txtName+","+txtPass+")";
这两句没必要了：
cmd.Parameters.Add("accountid", OracleType.VarChar, 32).Value = txtName;
            cmd.Parameters.Add("password", OracleType.VarChar, 8).Value = txtPass;

Answer 3

你去掉Try...catch看看是什么错误提示.

Answer 4

将参数值从 TextBox 转换到 String 失败

Answer 5

我把后面加上了text            cmd.Parameters.Add("user", OracleType.VarChar, 32).Value = txtName.Text;
            cmd.Parameters.Add("pass", OracleType.VarChar, 8).Value = txtPass.Text;错误变成了
ORA-01745: 无效的主机/赋值变量名

Answer 6

 cmd.Parameters.Add(":user", OracleType.Char).Value=txtName.Text;其中txtName为textbox 的id名

Answer 7

我原来写的OracleConnection conn = new OracleConnection(ConfigurationManager.ConnectionStrings["ConnStrOracle"].ConnectionString);//创建一个新连接
        conn.Open();
            OracleCommand cmd = new OracleCommand("INSERT INTO SITEAdmins (a_Id,a_Name,a_Ps,a_Power,a_Info,a_Block) VALUES (seq_admins.nextval,:a_Name,:a_Ps,:a_Power,:a_Info,:a_Block)", conn);
            cmd.Parameters.Add(":a_Name", OracleType.Char);
            cmd.Parameters.Add(":a_Ps", OracleType.Char);
            cmd.Parameters.Add(":a_Power", OracleType.Int16);
            cmd.Parameters.Add(":a_Info", OracleType.Char);
            cmd.Parameters.Add(":a_Block", OracleType.Int16);            cmd.Parameters[":a_Name"].Value = a_Name.Text;
            cmd.Parameters[":a_Ps"].Value = System.Web.Security.FormsAuthentication.HashPasswordForStoringInConfigFile(a_Ps.Text, "SHA1");
            cmd.Parameters[":a_Power"].Value = a_Power.Text;
            cmd.Parameters[":a_Info"].Value = a_Info.Text;
            cmd.Parameters[":a_Block"].Value = ListBox_blocks.SelectedItem.Value;            cmd.ExecuteNonQuery();
            cmd.Dispose();    
        

Answer 8

using (OracleCommand cmd = new OracleCommand())
            {
                using(OracleConnection connection = new OracleConnection(""))
                {
                connection.Open();
                cmd.Connection = connection;
                cmd.CommandType = CommandType.StoredProcedure;
                cmd.CommandText = "";
                
               cmd.Parameters.Add("","参数值");
               
                  using (OracleDataReader dr = cmd.ExecuteReader())
                {                }
                }
            }
看看Exception提示什么

Answer 9

提示 ORA-00942: 表或视图不存在