我给出的代码是--------
SqlConnection cn = new SqlConnection(@"Data Source=CHROMA\SQLEXPRESS;Initial Catalog=marksheet;Integrated Security=True");
cn.Open();
if (cn != null)
{
Console.WriteLine("Enter roll number");
string rollnum =Console.ReadLine();
Console.WriteLine("Enter name");
string name = Console.ReadLine();
Console.WriteLine("Enter gender");
string gen = Console.ReadLine();
Console.WriteLine("Enter DOB");
string DOB = Console.ReadLine();
Console.WriteLine("Enter father name");
string father = Console.ReadLine();
Console.WriteLine("Enter course");
string course = Console.ReadLine();
Console.WriteLine("Enter address");
string add = Console.ReadLine();
Console.WriteLine("Enter city");
string city = Console.ReadLine();
Console.WriteLine("Enter state");
string state = Console.ReadLine();
Console.WriteLine("Enter phone");
string phone = Console.ReadLine();
SqlCommand cmd = new SqlCommand("insert into student (rollno,name,gender,dob,fname,course,address,city,state,phone) values(@rollnum,@name,@gen,@DOB,@father,@course,@add,@city,@state,@phone,)",cn);
int row= cmd.ExecuteNonQuery();
if(row >0)
Console.WriteLine("Record inserted...");
cn.Close();
cn.Dispose();
}
else
Console.WriteLine("please open connection first!!");
最佳答案
您应该在创建命令之后和执行命令之前绑定(bind)您的参数。
例如:
cmd.Parameters.AddWithValue("@rollnum", rollnum);
(对所有其他参数执行相同操作)
关于c# - 为什么我尝试插入时出现 "Must declare the scalar variable "@rollnum""错误？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/28851075/