How to reload the data grid view during runtime c#

Question

I have reload the data for data grid view each time the form open.

 Student_DetailEntities db = new Student_DetailEntities();
 private void Form1_Load(object sender, EventArgs e)
 {
     db.StudentTables.Load();
     studentTableBindingSource.DataSource = db.StudentTables.Local;
 }

To ensure the data table is refresh and show in data grid view, I have tried this code

private void refreshToolStripMenuItem_Click(object sender, EventArgs e)
{
        Application.Restart();
        Environment.Exit(0);
}

and I try

private void refreshToolStripMenuItem_Click(object sender, EventArgs e)
{
        this.Controls.Clear();
        this.InitializeComponent();
}

But the data grid view still not reload. Each time I add new item, I need to close and open the form again for it to show in data grid view.

When I run the program, it will open form1 which contain the data grid view. The other form to add the new item is open only if the user click menu strip on form1 — irham dollah, Jun 26 '20 at 05:39
You should put the code of `Form1_Load` in `refreshToolStripMenuItem_Click` — Chetan, Jun 26 '20 at 05:39
I have tried it. But it just make it redundant. Another menu strip showing up under the previous menu strip. — irham dollah, Jun 26 '20 at 06:03

Selim Yildiz · Accepted Answer · 2020-06-26T06:56:14.820

2

You can use Shown event instead of Form.Load.So that you can reload the data each time the form is shown.

Student_DetailEntities db = new Student_DetailEntities();
private void Form1_Load(object sender, EventArgs e)
{
    Shown += Form1_Shown; 
}
 
 
private void Form1_Shown(object sender, EventArgs e) 
{
     db.StudentTables.Load();
     studentTableBindingSource.DataSource = db.StudentTables.Local;
}

See the detail of Form.Load, Form.Shown:

Order of events 'Form.Load', 'Form.Shown' and 'Form.Activated' in Windows Forms

@Caius Jard pointed out in comments, you can simply add Form.Shown event handler by "Form Designer --> Properties --> Events --> double click Shown then add code" instead of Shown += Form1_Shown; in Form1_Load that I mentioned above.

edited Jun 26 '20 at 06:56

answered Jun 26 '20 at 05:50

Selim Yildiz

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You are right. I have added your suggestion as well. Thanks for contribution – Selim Yildiz Jun 26 '20 at 06:57
I have added the `form_1_shown` and `form_1_load` as you show above. But when I do `application.restart`, it still not reload. The data grid view still showing same result. – irham dollah Jun 26 '20 at 07:28
You don't need to `Application.Restart` just try to use [`ShowDialog`](https://learn.microsoft.com/en-us/dotnet/api/system.windows.forms.form.showdialog) . – Selim Yildiz Jun 26 '20 at 07:38
Do u mean I use `Form f = new Form(); f.ShowDialog = DialogResult.OK; ` under `refreshToolStripMenuItem_Click`? – irham dollah Jun 26 '20 at 07:57
It depends what you want, You can use ShowDialog as `Form f = new Form(); f.ShowDialog();` or you can use `var f = new Form(); f.Show();`. https://stackoverflow.com/questions/3965043/how-to-open-a-new-form-from-another-form – Selim Yildiz Jun 26 '20 at 08:04
I think it is not working. It just pop up a new form, without reloading the grid view – irham dollah Jun 26 '20 at 09:13
Put debugger the entrance of `Form1_Shown`. See if method is fired when you open the form. – Selim Yildiz Jun 26 '20 at 09:55

How to reload the data grid view during runtime c#

1 Answers1