GUI Button OnMouseUp?

Archived Forums Questions & Answers
1

Hello, how can I make something happen when I release my mouse from my button?

This is my code in C# for the MouseDown:

if(GUI.Button(new Rect(0, ((Screen.height / 2) - 128), 256, 256), "", GUILeft)) {
     TurnLeftInputDown();
}

This is how it should act:
When you press the button it should run the “TurnleftInputDown” function 1 time, that’s why I don’t use RepeatButton.

And when you release the mouse, it should run another function.

Thanks, Andreas :slight_smile:

Actually GUI buttons execute on mouse up, so your real question should be how to execute a function on the mouse down.

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2 Answers

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You can test for the MouseDown event and test the position against the Rect you use for your button. Example:

private var rect = Rect(0, ((Screen.height / 2) - 128), 256, 256);

function OnGUI() {
	var e = Event.current;
	if ((e.type == EventType.MouseDown) && rect.Contains (Event.current.mousePosition))
		Debug.Log("Mouse Down");
	
	if(GUI.Button(rect, "Button")) {
     	Debug.Log("Mouse Up");
	}
  }

Thanks :), but I get one error. "The name e' does not exist in the current context", could it be that it's javascript and I'm using C#? And it's on this line: e = Event.current;` I have changed the rect to Rect rect = new Rect(0, ((Screen.height / 2) - 128), 256, 256); :-)

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I guess I was right, I changed e = Event.current; to Event e = Event.current; and now it works perfectly, thank you so much! :D :D :D :D

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The code should have been var e since it was Javascript. Event e is right for C#. And you need the 'new' again because you are writing in C# and I posted the example in Javascript.

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3

Does anyone know the answer in JavaScript. I’d like to hear it!