CODEWITHSUNDEEP
c#.netwinforms
mafap
mafap

Reputation: 391

Keeping values of secondary Form

I have two Forms: MainForm and OptionsForm, wich has a button (OK) that applies the some changes on MainForm. When I open the OptionsForm for the first time everything is ok, with the default values.

After I make some changes and and click OK the options are applied but when I open the OptionsForm for the second time, I wanted to hold the previous values, not the default ones like its happening.

OptionsForm is opened through MainForm like this

OptionsForm formOptions = new OptionsForm();
if (formOptions.ShowDialog(this) == DialogResult.OK)
{
    // etc..
}

//...

public string otherLabel
    {
        get { return formMainLabel.Text; }
        set { formMainLabel.Text = value; }
    }

In OptionsForm I have a NumericUpDown and want to hold its value

private MainForm mainForm = null;
public OptionsForm(Form callingForm)
{
    mainForm = callingForm as MainForm;
    InitializeComponent();
}
// ...
private void btnOK_Click(object sender, EventArgs e)
{
     this.mainForm.someLabel= someBox.Value.ToString(); // NumericUpDown
     this.mainForm.otherLabel = "abc";       //>>> Getting NullReferenceException
     this.Close();
}

Now I can hold my settings but I'm getting a NullReferenceException. I tryed this but it's still not working. Any sugestion?

Upvotes: 1

Views: 490

Answers (6)

Abdul Saleem
Abdul Saleem

Reputation: 10622

Seems like you are calling an instance of the options form from a click event like this:

button1_click(object sender, EventArgs e)
{
    OptionsForm optForm = new OptionsForm();
    optForm.showDialog();
}

You have to create variable for the options form class within the Mainform class, instantiate it there, or in the constructor, and only call the ShowDialog() or Show() method within the button_Click event. Like this:

partial class MainForm:Form
{
    OptionsForm optForm;
    ............
    ............
    public MainForm()    //Constructor
    {
        initialiseComponent();
        optForm = new OptionsForm();
        ........
    }

    .......

    private button1_Click(object sender, EventArgs e)
    {
        optForm.Show();    // or ShowDialog()
    }
}

and use this.hide() instead of this.close() in the options form.. or else the form gets disposed..

Upvotes: 0

Abdul Saleem
Abdul Saleem

Reputation: 10622

if you just want to keep only one numeric up/down control's value, then just pass it as a parameter in the constructor..

public FormOptions(MainForm -mainFrm, int curNumericValue)
{
    someBox.value = curNumericValue;
}

and instantiate the form with the value from the MainForm

private void button1_click(....)
{
    FormOptions formOptions = new FormOptions(this, Convert.toInt32(someLabel.text));
    formOptions.ShowDialog();
}

Upvotes: 1

Lee Harrison
Lee Harrison

Reputation: 2453

Why not just make use of the "Settings" that are available within the build properties, and WinForms itself? Just have your settings form populate itself with these variables, and "set" them when you click OK. Then have the MainForm refresh from these settings once the SettingsForm has fully closed. Easy, no mess, and no passing variables between forms.

Overview: http://msdn.microsoft.com/en-us/library/k4s6c3a0.aspx

SO Question covering Settings: Save Settings in a .NET Winforms Application

It also has the added bonus that the values will be saved between sessions of the application.

Upvotes: 0

Hans Passant
Hans Passant

Reputation: 942438

ShowDialog() was already made to support this. It is different from Show(), other than it being modal, it also prevents the form object from being disposed when the user closes it. So you can simply call ShowDialog() again, the controls keep their original values:

    private OptionsForm options = new OptionsForm();

    private void button1_Click(object sender, EventArgs e) {
        if (options.ShowDialog(this) == DialogResult.OK) {
            // etc..
        }
    }

    protected override void OnFormClosed(FormClosedEventArgs e) {
        options.Dispose();
        base.OnFormClosed(e);
    }

Upvotes: 5

aleppke
aleppke

Reputation: 496

One way to do it would be to just always keep a reference to your FormOptions and show the same instance rather than creating a new one every time. Or, if you don't want to do that, you can create an Options class that stores all of your options, which can then be stored and passed into any new instance you create thereafter. There's advantages and disadvantages to both so feel free to choose the option that best suits your needs.

public class MyOptions
{
    public String StringOption { get; set; }
    public int IntOption { get; set; }
}

Your FormOptions would then have a MyOptions property where you can set all of your options and retrieve them.

public class FormOptions : Form
{
    ...
    private MyOptions _options;
    public MyOptions Options 
    {
        get { return _options;}
        set
        {
            _options = value;
            // Set the Form's control values accordingly.
        }
    }
    ...
}

And finally, you would call it like so in your code:

    FormOptions optionsForm = new FormOptions();
    MyOptions savedOptions = new MyOptions(); // Probably don't want to create a new instance every time but I'm sure you get the idea here.
    optionsForm.Options = savedOptions;
    optionsForm.ShowDialog();

    // Get the new options after the form is closed.
    savedOptions = optionsForm.Options;

Upvotes: 1

dantix
dantix

Reputation: 745

Well, you should pass these values to OptionsForm, feel free to write custom constructor for OpptionsForm and call it when you need.

Upvotes: 1

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