Must declare the table variable @table
I'm a beginner in C# and SQL, i have this SQL insert statement that i want to perform. It asks for the table name among the other variables that i want to insert.
But when i run this console app i get this error :
Must declare the table variable @table
This is a part of the code :
StreamReader my_reader = getFile(args);
string CS = formCS();
try
{
using (SqlConnection con = new SqlConnection(CS))
{
SqlCommand com = new SqlCommand("insert into @table (time, date, pin) values (@time, @date, @pin)", con);
con.Open();
Console.WriteLine("Enter table name:");
Console.Write(">> ");
string tblname = Console.ReadLine();
com.Parameters.AddWithValue("@table", tblname);
string line = "";
int count = 0;
while ((line = my_reader.ReadLine()) != null)
{
Dictionary<string, string> result = extractData(line);
com.Parameters.AddWithValue("@time", result["regTime"]);
com.Parameters.AddWithValue("@date", result["regDate"]);
com.Parameters.AddWithValue("@pin", result["regPin"]);
count += com.ExecuteNonQuery();
com.Parameters.Clear();
}
Console.WriteLine("Recoreds added : {0}", count.ToString());
Console.WriteLine("Press Enter to exit.");
}
Console.ReadLine();
}
catch (SqlException ex)
{
Console.WriteLine(ex.Message);
}
catch (Exception ex)
{
Console.WriteLine(ex.Message);
}
You can't do this. You can't pass the table name as a parameter the way you did:
SqlCommand com = new SqlCommand("insert into @table ...");
...
com.Parameters.AddWithValue("@table", tblname);
You can do this instead:
Console.WriteLine("Enter table name:");
Console.Write(">> ");
string tblname = Console.ReadLine();
string sql = String.Format("insert into {0} (time, date, pin) values ... ", tblname);
SqlCommand com = new SqlCommand(sql, con);
...
The table name cannot be an input parameter in a sql query. However, you can always "prepare the sql string BEFORE passing it to the SqlCommand as follows:
var sqlString = string.Format("insert into {0} (time, date, pin) values (@time, @date, @pin)", tblname)
and then
SqlCommand com = new SqlCommand(sqlString);
...