Convert RectTransform rect to screen space

Question

I have a RectTransform that is the c$$anonymous$$ld of several other RectTransforms. Is there a way for me to convert the c$$anonymous$$ld RectTransform's rect property (w$$anonymous$$ch is in the local space of the transform) into global screen space (essentially pixel coordinates relative to a corner of the screen)?

Edit: My Canvas is set to Screen Space - Camera and is rendered by the main camera.

Answer 1

Answer by sourav13 · Jul 23, 2015 at 07:25 AM

You mean world space when you say global screen space .Change the render mode of canvas to world space first.

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rahulkk · Jul 23, 2015 at 03:16 PM 0

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I don't mean 3D world space. I need to get the 2D pixel coordinates of the rect relative to the corner of the screen.

Answer 2

Answer by Malkyne · Dec 13, 2015 at 08:14 PM

Try t$$anonymous$$s out:

     public static Rect RectTransformToScreenSpace(RectTransform transform)
     {
         Vector2 size = Vector2.Scale(transform.rect.size, transform.lossyScale);
         return new Rect((Vector2)transform.position - (size * 0.5f), size);
     }

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ZachariBarnes · Jun 10, 2018 at 04:33 PM 0

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This worked flawlessly for what I needed. Thank you

Zmeyk · Mar 23, 2021 at 11:21 AM 0

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Thanks, great solution. Just one correction in last string (because pivot may not be in the center):

 return new Rect((Vector2)transform.position - (size * transform.pivot), size);

Terrain22 · May 25, 2021 at 01:46 PM 0

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This doesn't work if the RectTransform has a negative scale on one axis (i assume the position probably went into the negatives, which it shouldn't have), which was very hard to figure out in my case for why just one of my buttons weren't working. To correct for that, you can use this code to calculate size (is there really no Vector3.Abs in Unity?)

 Vector2.Scale(transform.rect.size, new Vector3(Mathf.Abs(transform.lossyScale.x), Mathf.Abs(transform.lossyScale.y), Mathf.Abs(transform.lossyScale.z)))

iHaveReturnd · May 25, 2021 at 05:05 PM 0

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@Terrain22 I wouldn't add code to account for a negative scale on the UI, I think it would be much better practice to fix the UI scale as it can lead to a lot of problems down the road, and ones that are hard to troubleshoot if you don't fix it. If you have a really good reason for using negative UI scale I suppose that works, but would definitely correct it otherwise.

Terrain22 iHaveReturnd · May 25, 2021 at 08:40 PM 0

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I agree, it's probably a better idea to fix the scale in most cases. In my case, i was working on a project i didn't originally create the UI for, and it flipped a sprite using a negative scale - this was easily fixed since it was symmetric so i could just rotate it instead, but in other cases you might need a new sprite that's flipped instead of using the scale to flip it.

Answer 3

Answer by adamonline45 · Jan 15, 2016 at 11:20 PM

I had two challenges with @Malkyne's answer (w$$anonymous$$ch could be my own doing):

I didn't have a constructor for Rect w$$anonymous$$ch took two vectors
The position had to be converted to screen space (+y is down)

Here is what worked for me:

 public static Rect RectTransformToScreenSpace( RectTransform transform )
 {
     Vector2 size = Vector2.Scale( transform.rect.size, transform.lossyScale );
     return new Rect( transform.position.x, Screen.height - transform.position.y, size.x, size.y );
 }

Otherwise, thanks, @Malkyne!

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Answer 4

Answer by IVxIV · Sep 08, 2016 at 08:11 PM

Here's a slightly modified version of the above solution w$$anonymous$$ch also accounts for UI anchor positions (in case anyone else comes across t$$anonymous$$s topic like I did):

 public static Rect RectTransformToScreenSpace(RectTransform transform)
 {
     Vector2 size= Vector2.Scale(transform.rect.size, transform.lossyScale);
     float x= transform.position.x + transform.anchoredPosition.x;
     float y= Screen.height - transform.position.y - transform.anchoredPosition.y;
 
     return new Rect(x, y, size.x, size.y);
 }

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Answer 5

Although it's been a w$$anonymous$$le since t$$anonymous$$s question was active, I've stumbled across it and came up with an extended solution. Because I needed a similar function but the above solutions didn't take non-default pivots on a RectTransform into account, I've picked up adamonline45's solution and added a position s$$anonymous$$ft by the transforms pivot multiplied by it's screen size. The 1.0 - transform.pivot.y is done due to the flipped y-axis.

     public static Rect RectTransformToScreenSpace(RectTransform transform)
     {
         Vector2 size = Vector2.Scale(transform.rect.size, transform.lossyScale);
         Rect rect = new Rect(transform.position.x, Screen.height - transform.position.y, size.x, size.y);
         rect.x -= (transform.pivot.x * size.x);
         rect.y -= ((1.0f - transform.pivot.y) * size.y);
         return rect;
     }

I've tested it with some GUI.Box calls in an OnGUI() method, w$$anonymous$$ch worked fine as it overdraw the UI button I used as RectTransform.

Convert RectTransform rect to screen space

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