use*_*123 7 c# sql-server datagridview
我正在使用此视频尝试填充a的结果DataGridView
,并收到上述错误.下面的代码与此错误有关 - 我将值传递给存储过程,然后最终SELECT
返回表中的值DataGridView
.
SqlConnection con = new SqlConnection();
con.ConnectionString = "integrated security=SSPI;data source=SERV;" + "persist security info=False;initial catalog=DB";
con.Open();
SqlCommand select = new SqlCommand("SELECT * FROM Table");
SqlCommand enter = new SqlCommand("sp_Proc", con);
// Stored Procedure
enter.CommandType = CommandType.StoredProcedure;
enter.Parameters.Add(new SqlParameter("@vvalue", SqlDbType.VarChar)).Value = Convert.ToString(txt1.Text);
enter.Parameters.Add(new SqlParameter("@dvalue", SqlDbType.Decimal)).Value = Convert.ToDecimal(txt2.Text);
enter.ExecuteNonQuery();
// DataGrid returns the SELECT
SqlDataAdapter sadapt = new SqlDataAdapter(select);
sadapt.SelectCommand = select;
DataTable dtab = new DataTable();
sadapt.Fill(dtab); // generates the error
BindingSource b = new BindingSource();
b.DataSource = dtab;
dGrid.DataSource = b;
sadapt.Update(dtab);
con.Close();
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Joh*_*Woo 14
你没有传递connection
对象command
.试试这个,
SqlCommand select = new SqlCommand("SELECT * FROM Table", con);
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