SQL创建ACCESS表ACCESS是2003版的[解决办法]OleDbConnection cnnew OleDbConnection(连接字符串)OleDb
SQL创建ACCESS表
ACCESS是2003版的
[解决办法]
OleDbConnection cn=new OleDbConnection("连接字符串");
OleDbCommand cmd=new OleDbCommand("create table 测试(stuId varchar(10) primary key,name varchar(20))",cn);
cn.Open();
cmd.ExecuteNonQuery();
cn.Close();
