2011-08-17 69 views
0
protected void btnsubmit_Click(object sender, EventArgs e) 
    { 
     if (rbtmale.Checked) 
     { 
      gender = "male"; 
     } 
     else if(rbtfemale.Checked) 
     { 
      gender = "female"; 
     } 
     if (cbxsd.Checked) 
     { 
      designation = "sd"; 
     } 
     else if (cbxasd.Checked) 
     { 
      designation = "asd"; 
     } 
     else if (cbxtrainee.Checked) 
     { 
      designation = "trainee"; 
     } 
     if (cbxbsc.Checked) 
     { 
      qua = "BSc"; 
     } 
     else if (cbxbca.Checked) 
     { 
      qua = "BCA"; 
     } 
     else if (cbxbtech.Checked) 
     { 
      qua = "BTech"; 
     } 
     else if (cbxmca.Checked) 
     { 
      qua = "MCA"; 
     } 
     else if (cbxmba.Checked) 
     { 
      qua = "MBA"; 
     } 



     SqlConnection con = new SqlConnection(s1); 
     SqlCommand cmd = new SqlCommand("sp_2", con); 
     cmd.CommandType = CommandType.StoredProcedure; 
     con.Open(); 
     cmd.Parameters.AddWithValue("@id", txtid.Text.Trim()); 
     cmd.Parameters.AddWithValue("@name",txtname.Text.Trim()); 
     cmd.Parameters.AddWithValue("@age",txtage.Text.Trim()); 
     cmd.Parameters.AddWithValue("@sex",gender); 
     cmd.Parameters.AddWithValue("@addr",txtaddr.Text.Trim()); 
     cmd.Parameters.AddWithValue("@hq",qua); 
     cmd.Parameters.AddWithValue("@company",ddlcompany.SelectedItem); 
     cmd.Parameters.AddWithValue("@desig",designation); 
     cmd.Parameters.AddWithValue("@mob",txtmob.Text.Trim()); 
     cmd.Parameters.AddWithValue("@email", txtemail.Text.Trim()); 
     cmd.Parameters.Add("@msg", SqlDbType.VarChar, 20).Direction = ParameterDirection.Output; 
     cmd.ExecuteNonQuery(); 
     //if (i > 0) 
     //{ 
     // Label1.Text = "successfully inserted"; 
     //} 
     //else 
     //{ 
     // Label1.Text = "failed to insert"; 
     //} 
     string s2 = cmd.Parameters["@msg"].Value.ToString(); 
     Response.Write(s2); 
     con.Close(); 
    } 
+0

欢迎SO!考虑正确地格式化您的问题以提出具体的问题,并将代码仅限于有问题的部分。除非你改善这个问题,否则它肯定会被标记。 – Mrchief

回答

9

我想我看到的问题:

cmd.Parameters.AddWithValue("@company",ddlcompany.SelectedItem); 

应该是ddlcompany.SelectedIndexddlcompany.SelectedValue ...

ddlcompany.SelectedItem引用ListItem本身,而不是一个值。

5

只要改变这个下面的代码行

cmd.Parameters.AddWithValue("@company",ddlcompany.SelectedItem); 

cmd.Parameters.AddWithValue("@company",ddlcompany.SelectedItem.ToString()); 

这将工作的SelectedItem需要字符串类型转换

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