Widget issues

[Deep Slayer 7]

Hi all,
 

How do I avoid getting a null from a widget that isn't there?


So what I'm trying to do is check if a window is open.

But if the screen that I'm looking for is not open and I go to check if the Widget is not null.. it still throws a null? - because obviously the widget isn't open.

 

if(Widgets.getWidget(number1).getChild(number2)!=null)

maybe My logic is wrong? 

 

What I'm trying to do is check if a screen is open when it is not open. Does that make sense?

I'm looking for a way to do this ^^

 

Edited March 7, 2021 by Deep Slayer

[RetroBot 35]

WidgetChild example = Widgets.getWidgetChild(1, 1);

if (example != null && example.isVisible()) {
  //do stuff
} else {
  // open window
}

 

[Deep Slayer 7]

Hey RetroBot,

Thanks for your reply. I've tried this to check if its visible and != null. 

It still throws a Null pointer exception if I try to find the widget and it doesn't exist.

I'm not sure what the logic is supposed to be

[Deep Slayer 7]

if(Widgets.getWidget(446).getChild(35)!=null && Widgets.getWidget(446).getChild(35).isVisible()) {
    

}

this will throw an error if it can't find the widget - My thinking is that it should just skip this if statement if it cant find the widget?

Edited March 7, 2021 by Deep Slayer

[AsBakedAsCake 199]

2 hours ago, Deep Slayer said:

if(Widgets.getWidget(446).getChild(35)!=null && Widgets.getWidget(446).getChild(35).isVisible()) {
    

}

this will throw an error if it can't find the widget - My thinking is that it should just skip this if statement if it cant find the widget?

You're not doing something right and you probably have the wrong widget. You should do this like so:

<

WidgetChild randomWidg = Widgets.getWidgetChild(0,0,0);
if(randomWidg != null){
    log("Interact with widget.");
    randomWidg.interact();
}

Also use the game explorer to make sure the widget you're using has the action you need.

Edited March 7, 2021 by AsBakedAsCake

[Pseudo 178]

Try null checking the parent first, see if that works. Like so:

 

[code]

WidgetChild ourWidgetChild= Widgets.getWidget(446) != null ? Widgets.getWidgetChild(446,35) : null;
if (ourWidgetChild != null && ourWidgetChild.isVisible()) {

//do shit

}

/code]

[Deep Slayer 7]

Thanks for the help!

Issue has been solved. I have been using this guide to use my widgets: 

I think it may be out of date because in chapter 2, he explains to use this when using your widgets:

if (getWidgets().getWidget(parent).getChild(child) !=null)

obviously this syntax didn't work for what I am trying to do. Not sure if I missed something. But using Widgets.getWidgetChild(id's) seems
to have fixed this null error when the widget is not found.

[Libra12 2]

On 3/8/2021 at 8:29 PM, Deep Slayer said:

Thanks for the help!

Issue has been solved. I have been using this guide to use my widgets: 

I think it may be out of date because in chapter 2, he explains to use this when using your widgets:

if (getWidgets().getWidget(parent).getChild(child) !=null)

obviously this syntax didn't work for what I am trying to do. Not sure if I missed something. But using Widgets.getWidgetChild(id's) seems
to have fixed this null error when the widget is not found.

Thanks, i had the same problem, life saver!