I want to show a picture when app is opened

i have created an app & i want to show a picture when app is opened but initialize event is already used & i have created app with virtual screens so i have to use initialize event only & i know that when i return back to the real screen that picture will be shown that i don't want.. so is there any solution for this?

Please explain that in more detail.
Also show some screenshots of the Designer settings and post the blocks.

i m creating an app on Woman safety which was crashing earlier , so as suggested by @ChrisWard i made the app again with virtual screens & now i want to show a picture/ icon just like as we start an app like whatsapp or anything else so same i want to show & im not able to do that as the initialize block has already been used for other purpose

:point_up_2: the blocks

The rubber stamp is strong in this one.

sorry to say but im unable to understand

i think that i need to repeat myself in this case.. its complex but i will surely look over this :slightly_smiling_face:
thanks & regards

is there any solution for this?

Your app has lots of issues Neelam. You want to display an image when the app is opened. Use an Image control in your Screen1.Initialization event handler. Use set Image1.Picture to setting that to your startingImage.png or set your Screen1.BackgroundImage

The initialize block can be used for multiple purposes.

Your Blocks show two instances of screen1.Initialization. That is not allowed. Just add the Image1.Picture to blocks and code to display the arrangement showing your image (and at the same time hiding your other virtual screens.

Using the LocationSensor.LocationChanged event handler to reset the Texting message every time the LocationSensor achieves a satellite fix is awkward. You have a lot of redundant code. Be aware that unless you save values in your TinyDB for your Labels, the labels will be empty the first time the app is run.

thank you SteveJG ! :slightly_smiling_face:


You have moved over to Virtual Screens - correct? If so, is that why your button 1 click block (for example) is hiding components (visible = false) ?

Hi Neelam

Every component on an Arrangement is displayed/not displayed according to the display state of the Arrangement.

Look at this (click on the image for a zoomed view):

...If you want to show a "splash screen" on App Start, it can be on a vertical arrangement, timed to undisplay as required :slight_smile:

ok thanks @ChrisWard :slightly_smiling_face:

ya, you are totally correct. i am using button 1 click block (for example) is hiding components (visible = false) becoz i want to go back.. basically its working like back button...

So, as per my example, there is no need to hide the components, just the arrangement they belong to. Saves a lot of work and a lot of blocks.

When you click a button to show a different Virtual Screen, you don't need to know which VS is currently displayed, just set them all as visible = false and then set the one you want to see as visible = true, as per my example. It all happens in microseconds.

well.. i will be surely trying this but there's a prob. in spinner... whenever i select anything from that it opens the thing which we selected on the same screen without closing the previous virtual screen

Did you tell the Spinner that after it selects, it should close the previous virtual screen?
You perhaps need to do something like this spinnerSelectThenCloseVirtualScreen but will need to adapt what you place in the AfterSelecting event handler for your code. You might want to make another virtual screen visible at the same time (by setting that screen's Visible property to true).

yes ... i have written a code for that too..