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每次我在这里询问有关数据库的问题时，我都听说过参数化查询。看起来我没有使用参数化查询，我的代码可能会受到 SQL 注入的影响。所以这是我的代码：

public void CreateStudent(int ID, String status, String email, String firstName,     String lastName, String password, String level, String program)
{
  SqlConnection con = new SqlConnection(GetConnectionString());

  string query1 = "insert into StudentTable(Name,Surname,ID,email,level,program,status,password,Type) values ("
   + "'" + firstName + "'" + "," + "'" + lastName + "'" + ","
   + "'" + ID + "'" + "," + "'" + email + "'" + "," + "'" + level + "'" + "," + "'" + program + "'" + "," + "'" + status + "'"
   + "," + "'" + password + "'" + "," + "'" + "Student" + "'" + ")";

  SqlCommand command = new SqlCommand(query1,con);

  int result;
  con.Open();
  result = command.ExecuteNonQuery();
  con.Close();
}

这是我尝试过的：

SqlConnection con = new SqlConnection(GetConnectionString());

string query1 = "insert into StudentTable(Name,Surname,ID,email,level,program,status,password,Type) values(@firstName,@lastName,@ID,@email,@level,@program,@status,@password,Student)";

SqlCommand command = new SqlCommand(query1,con);

command.Parameters.AddWithValue("@firstName", firstName);
command.Parameters.AddWithValue("@lastName", lastName);
command.Parameters.AddWithValue("@ID", ID);
command.Parameters.AddWithValue("@email", email);
command.Parameters.AddWithValue("@level", level);
command.Parameters.AddWithValue("@program", program);
command.Parameters.AddWithValue("@status", status);
command.Parameters.AddWithValue("@password", password);

int result;
con.Open();
result = command.ExecuteNonQuery();
con.Close();

这给出了一个错误，指出 Student 是一个无效的列名。实际上，这里我尝试使用“Student”作为要添加到列 Type 的字符串值。有人可以将此查询编写为参数化查询以便我理解吗？

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2 回答 2

3

在这种情况下，它应该是'Student'

SqlConnection con = new SqlConnection(GetConnectionString());


string query1 = "insert into StudentTable(Name,Surname,ID,email,level,program,status,password,Type) values(@firstName,@lastName,@ID,@email,@level,@program,@status,@password,'Student')";


SqlCommand command = new SqlCommand(query1,con);

command.Parameters.AddWithValue("@firstName", firstName);
command.Parameters.AddWithValue("@lastName", lastName);
command.Parameters.AddWithValue("@ID", ID);
command.Parameters.AddWithValue("@email", email);
command.Parameters.AddWithValue("@level", level);
command.Parameters.AddWithValue("@program", program);
command.Parameters.AddWithValue("@status", status);
command.Parameters.AddWithValue("@password", password);

int result;
con.Open();
result = command.ExecuteNonQuery();
con.Close();
于 2013-06-01T22:25:30.347 回答
0

检查此链接

    public void CreateStudent(int ID, String status, String email, String firstName, String lastName, String password, String level, String program)
    {
        SqlConnection con = new SqlConnection(GetConnectionString());

        using (
            SqlCommand command =
                new SqlCommand(
                    @"insert into StudentTable(Name,Surname,ID,email,level,program,status,password,Type) values 
                    (@name, @surname, @id, @email, @level, @program, @status,@password,'Student')",
                    con))
        {
            //
            // Add new SqlParameter to the command.
            //
            command.Parameters.Add(new SqlParameter("name", firstName));
            command.Parameters.Add(new SqlParameter("surname", lastName));
            command.Parameters.Add(new SqlParameter("id", ID));
            command.Parameters.Add(new SqlParameter("email", email));
            command.Parameters.Add(new SqlParameter("level", level));
            command.Parameters.Add(new SqlParameter("program", program));
            command.Parameters.Add(new SqlParameter("status", status));

            int result;
            con.Open();
            result = command.ExecuteNonQuery();
            con.Close();
        }
    }
于 2013-06-01T22:30:00.047 回答