Question

我已加入两个表值并将这些值插入另一个表中，但插入不起作用。我正在使用Access数据库，我使用以下代码：

 string query = "select t2.date,t1.FlightNo,t1.Dept_Time,t1.Arr_Time,t1.Route,t1.[Destination 1],t1.[Destination 2],t1.[Destination 3],t1.[Destination 4] From [FlightNumber]as t1 inner join [schedule]as t2 on t1.FlightNo=t2.FlightNo";

 OleDbCommand cmd = new OleDbCommand(query, con);

 OleDbDataAdapter adp = new OleDbDataAdapter(cmd);
 dt1 = new DataTable();
 adp.Fill(dt1);
 con.Open();
 string query2 = "Insert into Monday (date,[Flight_no],[Dept_time],[Arr_time],Route,[dest_1],dest2,dest3,dest4)values(@date,@flight,@dept,@arr,@route,@dest1,@dest2,@dest3,@dest4)";
 for (int i = 0; i < dt1.Rows.Count; i++)
 {
     OleDbCommand cmd1 = new OleDbCommand(query2, con);

     cmd.Parameters.AddWithValue("@date", SqlDbType.NVarChar).Value = dt1.Rows[i]["date"].ToString();
     cmd.Parameters.AddWithValue("@flight", dt1.Rows[i]["FlightNo"]);
     cmd.Parameters.AddWithValue("@dept", dt1.Rows[i]["Dept_Time"]);
     cmd.Parameters.AddWithValue("@arr", dt1.Rows[i]["Arr_Time"]);
     cmd.Parameters.AddWithValue("@route", dt1.Rows[i]["Route"]);
     cmd.Parameters.AddWithValue("@dest1", dt1.Rows[i]["Destination 1"]);
     cmd.Parameters.AddWithValue("@dest2", dt1.Rows[i]["Destination 2"]);
     cmd.Parameters.AddWithValue("@dest3", dt1.Rows[i]["Destination 3"]);
     cmd.Parameters.AddWithValue("@dest4", dt1.Rows[i]["Destination 4"]);

     cmd1.ExecuteNonQuery();
     con.Close();
     MessageBox.Show("successfully inserted");            
 }

Answer 1

DATE是Access SQL中的reserved word，因此您需要在INSERT命令中将该列名包装在方括号中（就像使用[Flight_no]，[Dept_time]等一样）

string query2 = "Insert into Monday ([date], ...

Answer 2

除了Gord提到的保留字Date之外，如果你不继续重建命令和参数，你可能会在性能上更好，特别是如果你有很多想要插入的记录。相反，做一些像......

string query2 = "Insert into Monday (date,[Flight_no],[Dept_time],[Arr_time],Route,[dest_1],dest2,dest3,dest4)values(@date,@flight,@dept,@arr,@route,@dest1,@dest2,@dest3,@dest4)";

//预先构建命令和参数ONCE
OleDbCommand cmd1 = new OleDbCommand（query2，con）;

cmd.Parameters.AddWithValue（“@ date”，dt1.Rows [i] [“date”]。ToString（））; cmd.Parameters.AddWithValue（“@ flight”，dt1.Rows [i] [“FlightNo”]）; cmd.Parameters.AddWithValue（“@ dept”，dt1.Rows [i] [“Dept_Time”]）; cmd.Parameters.AddWithValue（“@ arr”，dt1.Rows [i] [“Arr_Time”]）; cmd.Parameters.AddWithValue（“@ route”，dt1.Rows [i] [“Route”]）; cmd.Parameters.AddWithValue（“@ dest1”，dt1.Rows [i] [“Destination 1”]）; cmd.Parameters.AddWithValue（“@ dest2”，dt1.Rows [i] [“Destination 2”]）; cmd.Parameters.AddWithValue（“@ dest3”，dt1.Rows [i] [“Destination 3”]）; cmd.Parameters.AddWithValue（“@ dest4”，dt1.Rows [i] [“Destination 4”]）;

 for (int i = 0; i < dt1.Rows.Count; i++)
 {
     cmd.Parameters[0].Value = dt1.Rows[i]["date"].ToString();
     cmd.Parameters[1].Value = dt1.Rows[i]["FlightNo"]);
     cmd.Parameters[2].Value = dt1.Rows[i]["Dept_Time"]);
     cmd.Parameters[3].Value = dt1.Rows[i]["Arr_Time"]);
     cmd.Parameters[4].Value = dt1.Rows[i]["Route"]);
     cmd.Parameters[5].Value = dt1.Rows[i]["Destination 1"]);
     cmd.Parameters[6].Value = dt1.Rows[i]["Destination 2"]);
     cmd.Parameters[7].Value = dt1.Rows[i]["Destination 3"]);
     cmd.Parameters[8].Value = dt1.Rows[i]["Destination 4"]);

     cmd1.ExecuteNonQuery();
 }

 con.Close();
 MessageBox.Show("successfully inserted");

从C＃应用程序插入Access数据库失败

2 个答案: