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Question by rahulkk · Jul 23, 2015 at 04:03 AM · parentrecttransformparent-childscreenspacelocalspace

Convert RectTransform rect to screen space

I have a RectTransform that is the child of several other RectTransforms. Is there a way for me to convert the child RectTransform's rect property (which 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.

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avatar image iHaveReturnd · Jul 23, 2015 at 03:55 PM 0
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Is this what you are looking for?

http://answers.unity3d.com/questions/826851/how-to-get-screen-position-of-a-recttransform-when.html

avatar image rahulkk · Jul 23, 2015 at 04:26 PM 0
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I have looked at that, but it doesn't seem to work. Here is the code I wrote, please let me know if I should change anything. $$anonymous$$y Canvas is set to Screen Space - Camera and is rendered by the main camera. This script is attached to the RectTransform whose rect I need to find in screen coordinates:

 Vector2 screenPos = RectTransformUtility.WorldToScreenPoint(Camera.main, transform.position);
 print(screenPos);

However, while my RectTransform is entirely on screen (its pivot point and anchors are on screen too), the script outputs:

(-116.3, -429.3), which is not what I would expect. $$anonymous$$aybe the RectTransform position isn't correct, but I don't know to fix that.

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Answer by Dowen · May 17, 2018 at 07:42 PM

In 2018, What I have used is RectTransformUtility, by Unity Official Support.

 RectTransform yourRectTransform = yourObject.getComponent<RectTransform>();
 Canvas root = yourRectTransform.transform.transform.root.GetComponent<Canvas>();
 Rect rect = RectTransformUtility.PixelAdjustRect(yourRectTransform, root);

That's it.

Be Noticed: The origin is not what you thought (My guess is 0,0 in the center of screen), you might see the negative x and y, but Width and Height are simply correct in Screen Space.

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avatar image Carocrazy132 · Feb 12, 2019 at 07:05 PM 0
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Why are you getting the transform of a transform of a transform? This answer worked less than almost any other answer here for me.

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Answer by al-gco · Apr 04, 2018 at 10:49 PM

None of the answers here worked for me, and it keeps being first in my Google searches, so hopefully this will be helpful to some future similar soul. It works for my use case as far as I've tested it.

The returned Rect is in screen coordinates, which is to say 0,0 is at the top left corner and y+ is down.

     public static Rect RectTransformToScreenSpace(RectTransform transform)
     {
         Vector2 size = Vector2.Scale(transform.rect.size, transform.lossyScale);
 
         float newRectX = transform.position.x + (Screen.width / 2) - (size.x * (1 - transform.pivot.x));
         float newRectY = Mathf.Abs(transform.position.y - (Screen.height / 2)) - (size.y * (1 - transform.pivot.y));
 
         return new Rect(newRectX, newRectY, size.x, size.y);
     }
 
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Answer by Tobias-Pott · Nov 02, 2016 at 03:28 PM

Although it's been a while since this 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 shift 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, which worked fine as it overdraw the UI button I used as RectTransform.

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

Here's a slightly modified version of the above solution which also accounts for UI anchor positions (in case anyone else comes across this 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 by adamonline45 · Jan 15, 2016 at 11:20 PM

I had two challenges with @Malkyne's answer (which could be my own doing):

  • I didn't have a constructor for Rect which 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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