【发布时间】:2020-05-06 00:16:07
【问题描述】:
Using this earlier question I need a bit of help
使用上面链接中的第二个答案,我必须为 MySQL 更新它
private void btLogin_Click(object sender, EventArgs e)
{
string connectionString;
connectionString = "SERVER=" + server + ";" + "DATABASE=" + database + ";" + "UID=" + uid + ";" + "PASSWORD=" + password + ";";
connection = new MySqlConnection(connectionString);
using (var con = new MySqlConnection(connectionString));
{
using (var command = new MySqlCommand(connection = con))
{
con.Open();
command.CommandText = @"SELECT level FROM userTable WHERE user=@username, password=@password";
command.Parameters.AddWithValue("@username", lbUser.Text);
command.Parameters.AddWithValue("@password", tbPassword.Text);
var strLevel = command.ExecuteScalar();
if (strLevel == DBNull.Value || strLevel == null)
{
MessageBox.Show("Invalid username or password");
return;
}
else
{
MessageBox.Show("Successfully login");
Hide(); // hide this form and show another form
}
}
}
}
一切看起来都不错,但是这个
using (var con = new MySqlConnection(connectionString));
{
using (var command = new MySqlCommand(connection = con))
{
con.Open();
它说 con 不存在。不知道用那个井看问题。
【问题讨论】:
-
我找到了 con 没有退出的原因;有一个 ;第一次使用后。但是我现在有了这个:'using (var command = new MySqlCommand(connection = con))' Error CS1503 Argument 1: cannot convert from 'MySql.Data.MySqlClient.MySqlConnection' to 'string' DPinBox2 C:\Users\de507\源\repos\DPinBox2\DPinBox2\LoginForm.cs 76 活动
-
试试
using (var command = new MySqlCommand(con))。我认为您不再需要connection变量了。