Notify popup has been made

salman mohammad

Notify popup has been made.But the thing is if any changes made in database simultaneously the updated message is not being displayed in the notification popup.

Can anyone help me get out of it?

using System;

using System.Collections.Generic;

using System.ComponentModel;

using System.Data;

using System.Drawing;

using System.Linq;

using System.Text;

using System.Threading.Tasks;

using System.Windows.Forms;

using Tulpep.NotificationWindow;

using System.Data.SqlClient;

using System.Collections;

namespace NotificationPopup

{

public partial class NotificationPopup : Form

{

SqlCommand cmd;

SqlConnection con;

public NotificationPopup()

{

InitializeComponent();

}

PopupNotifier popup;

private void btnShow_Click(object sender, EventArgs e)

{

popup = new PopupNotifier();

popup.Image = Properties.Resources.notification;

popup.TitleText = "HELLO";

popup.ContentText = "WELCOME TO INDIA !";

popup.Popup();

}

ArrayList al;

int j;

private void timer1_Tick(object sender, EventArgs e)

{

if (j == 0)

{

con = new SqlConnection("Data Source=SALMAN\\SQLExpress;Database=MsDemo;Integrated Security=true");

cmd = new SqlCommand("SELECT Output FROM DemoOnPopup", con);

con.Open();

SqlDataReader dr = cmd.ExecuteReader();

al = new ArrayList();

while (dr.Read())

{

al.Add(dr[0].ToString());

i++;

}

popup = new PopupNotifier();

popup.Image = Properties.Resources.notification;

popup.BodyColor = Color.Yellow;

popup.ContentColor = Color.Black;

if (j

popup.ContentText = al[j].ToString();

popup.Popup();

if (timer1.Interval==500)

popup.Hide();

j++;

if (j == i)

timer1.Stop();

}

int i;

private void NotificationPopup_Load(object sender, EventArgs e)

{

}

6 Replies

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Amit Gupta8y ago
Accepted answer
Your code is good and independent, explain your issue
salman mohammad8y ago
There's no issue with the debugging part as said earlier.
I need a database monitor which monitors the changes made in the database and displays the updated message.
Amit Gupta8y ago
According to your code, it seems that you are calling your database in half a second which cause huge degradation in performance.
salman mohammad8y ago
There's no issue with debugging part .
It works fine.
Whatever the values stored (in the column) in the database are being displayed on the popup window.
My only problem is if any changes or updates made in the database it should simultaneously show the changed or updated message.
In a simpler words it should act as a monitor.
Amit Gupta8y ago
You have to check this by debugging your code, I beleive that the timer.stop was the root problem, when its stop then you can't able to show popup
salman mohammad8y ago
My problem is,if any changes made in the databse it's not reflecting in the popup window.
So,I donot know further proceeding of it and the logic to be written.