у меня есть 2 формы, они обе имеют label1, form1 и form2 label1 должны показывать один и тот же вывод, когда я нажимаю кнопку на form1, label1 в form1 изменится так, что label1 в form2
Форма 1
SqlDataReader reader = cmdnext.ExecuteReader();
while (reader.Read())
{
label1.Text = reader[0].ToString();
break;
}
sqlcon.Close();
Lobbypage lp = new Lobbypage(label1.Text);
}
Form2
public Lobbypage(string labelText)
{
InitializeComponent();
label1.Text = labelText;
}
кнопка пропуска
private void button1_Click(object sender, EventArgs e)
{
sqlcon.Open();
SqlCommand cmdcurrent = sqlcon.CreateCommand();
cmdcurrent.CommandType = CommandType.Text;
cmdcurrent.Parameters.Add("@Title", SqlDbType.VarChar).Value = title;
cmdcurrent.CommandText = "update tblQLCashier set status = 'missing' where queID = (select min(queID) from tblQLCashier where status = 'On process' and department = @Title)";
cmdcurrent.ExecuteNonQuery();
SqlCommand cmdnext = sqlcon.CreateCommand();
cmdnext.CommandType = CommandType.Text;
cmdnext.Parameters.Add("@Title", SqlDbType.VarChar).Value = title;
cmdnext.CommandText = "update tblQLCashier set status = 'On process' , department = @Title where queID = (select min(queID) from tblQLCashier where status = 'Pending' and department ='')";
cmdnext.ExecuteNonQuery();
sqlcon.Close();
sqlcon.Open();
cmdnext.CommandText = "select queID from tblQLCashier where queID = (select min(queID) from tblQLCashier where department in ('', @Title) and status in ('Pending', 'On process'))";
SqlDataReader reader = cmdnext.ExecuteReader();
while (reader.Read())
{
label1.Text = reader[0].ToString();
break;
}
sqlcon.Close();
Lobbypage lp = new Lobbypage(label1);
}
оба label1 в form1 и form2 должны одновременно обновляться и иметь одинаковый вывод
![enter image description here](https://i.stack.imgur.com/pnaGk.png)