CODEWITHSUNDEEP
c#imagerandomnumbersjpeg
Adi
Adi

Reputation: 921

How to Generate random images through numbers c#?

I am making an app in which 4 chits are to be thrown randomly. I am done with generating random numbers by clicking on a button random numbers are generated between 1 to 4.

My question is how do I associate the 4 images with those numbers? Suppose if 1 comes randomly - then 1.jpg should be displayed. My programming language is c# and working in Visual studio 2012.

public sealed partial class MainPage : Page

{

   Random rand = new Random();

 public MainPage()
    {

        this.InitializeComponent();
    }

    protected override void OnNavigatedTo(NavigationEventArgs e)
    {
    }

    private void Button_Click_1(object sender, RoutedEventArgs e)
    {
        int num=rand.Next(1,5);
        textblock1.Text=num.ToString();
    }}}

i m showing the numbers in a textblock .now i have 4 images which i have to display randomly.,my question is how and where to display them .

Upvotes: 1

Views: 10069

Answers (5)

Chris Minoko
Chris Minoko

Reputation: 1

Random rand = new Random(); int randomPosition = rand.Next(0, 9);

            lblEnglish.Text = sw.EnglishHeartSentence[randomPosition];
            lblShona.Text = sw.shonaSentence[randomPosition];
            switch (randomPosition)
            {
                case 0:
                    mainPicBox.Image = Funda_isizulu_.Properties.Resources.diarrhea_900x600px;
                    break;
                case 1:
                    mainPicBox.Image = Funda_isizulu_.Properties.Resources.hunger1; ;
                    break;
                case 2:
                    mainPicBox.Image = Funda_isizulu_.Properties.Resources.stomachache1;
                    break;
                case 3:
                    mainPicBox.Image = Funda_isizulu_.Properties.Resources.full_clipart_1;
                    break;
                case 4:
                    mainPicBox.Image = Funda_isizulu_.Properties.Resources.Top_10_Foods_To_Relieve_Menstrual_Cramps_1;
                    break;
                case 5:
                    mainPicBox.Image = Funda_isizulu_.Properties.Resources.preg;
                    break;
                case 6:
                    mainPicBox.Image = Funda_isizulu_.Properties.Resources.belly;
                    break;
                case 7:
                    mainPicBox.Image = Funda_isizulu_.Properties.Resources.a_man_with_visible_spine_holds_his_back;
                    break;
                case 8:
                    mainPicBox.Image = Funda_isizulu_.Properties.Resources.bell;
                    break;
                case 9:
                    mainPicBox.Image = Funda_isizulu_.Properties.Resources.coffee_growl_905;
                    break;

            }

Upvotes: 0

Adi
Adi

Reputation: 921

Done used this method 

private void Button_Click_1(object sender, RoutedEventArgs e)

    {

        Random rand = new Random();
        int dice = rand.Next(1, 5);

        switch (dice)
        {
            case 1:

                Image1.Source = new BitmapImage(new Uri(@"ms-appx:///Assets/1.png"));

                break;
            case 2:

                two.Source = new BitmapImage(new Uri(@"ms-appx:///Assets/2.png"));
                break;
            case 3:

                three.Source = new BitmapImage(new Uri(@"ms-appx:///Assets/3.png"));
                break;
            case 4:

                four.Source = new BitmapImage(new Uri(@"ms-appx:///Assets/4.png"));
                break;

        }

    }for bitmap image refrence will be "using Windows.UI.Xaml.Media.Imaging;"

Upvotes: 3

Jodrell
Jodrell

Reputation: 35726

er,

switch (number)
{
    case 1:
        return "1.jpg";
    case 2:
        return "2.jpg";
    case 3:
        return "3.jpg";
    case 4:
        return "4.jpg";
    default:
        throw ArgumentException("number");
}

it all depends really,

how about,

private static readonly WhateverTypeMyImagesNeedToBe[] images =
    {
        new WhateverTypeMyImagesNeedToBe("1.jpg"),
        new WhateverTypeMyImagesNeedToBe("2.jpg"),
        new WhateverTypeMyImagesNeedToBe("3.jpg"),
        new WhateverTypeMyImagesNeedToBe("4.jpg")
    }

private static WhateverTypeMyImagesNeedToBe GetRandomImage()
{
    return images[new Random().Next(0, images.Length)];
}

Upvotes: -1

Omar
Omar

Reputation: 16623

What about:

Random rand = new Random();

Bitmap GetRandomImage(){
    Bitmap bmp;
    int n = rand.Next(0, 4);

    if(n == 0)
       bmp = new Bitmap("mypath1");
    else if(n == 1)
       bmp = new Bitmap("mypath2");
    else if(n == 2)
       bmp = new Bitmap("mypath3");
    else
       bmp = new Bitmap("mypath4");

    return bmp;
}

Or you can rename your files and use something like:

Random rand = new Random();

Bitmap GetRandomImage(){
    int n = rand.Next(0, 4);

    return new Bitmap(string.Format("mypath\\{0}.jpg", n));
}

Upvotes: 0

L-Four
L-Four

Reputation: 13541

You could just substitute the generated number to a file name like: 

string.Format("{0}.jpg", generatedNumber);

Very little code, no 'if' or 'switch' needed.

I would also make a separate component (I)FileNameResolver that contains the logic to determine the correct file name and error handling; and use that one in your code.

Upvotes: 1

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