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Question by el_kloklo · Jun 06, 2013 at 08:25 PM · c#collisiongameobjectdetectionrigibody

detecting rigidbody collsion with no extra script

Hello All.

I am interested into accessing collision status of rigidbody with no extra script attached onto it, and which is not working with event.

I am currently having something theorically like this

 public class MyClass: MonoBehaviour {
     Rigidbody PositionReferenceA;
     Rigidbody PositionReferenceB;
     void Start () {
         PositionReferenceA = (Rigidbody)(new GameObject()).AddComponent("Rigidbody ");
         PositionReferenceB = (Rigidbody)(new GameObject()).AddComponent("Rigidbody ");
     }
 }

And I would like to have something like this in my class :

 void Update() {
     if(PositionReferenceA.IscurrentlyColliding()==true)
         //do stuff here
 }

Is there anyway to do this? I don't see any accesser like this for the rigidbody class, but maybe there is a tricky way to do this?

Thanks for your time.

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avatar image el_kloklo · Jun 12, 2013 at 04:11 PM 0
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To myself, and anyone confronted to the same problematic : It is necessary to have a counter of collision, as there could be multiple collision, for example :

 void OnTriggerEnter(Collider collidingObject) {
     Debug.Log (this.name+" object has collision with " +collidingObject.gameObject.name);
         CollisionQtty++;
 }
 
 void OnTriggerExit(Collider collidingObject) {
     Debug.Log (this.name+" object lost collision with " +collidingObject.gameObject.name);
         CollisionQtty--;
 }

In this case, it is just necessary to test if CollisionQtty is equal to 0.

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Answer by Benproductions1 · Jun 07, 2013 at 12:25 AM

Hello,

Other than writing your own rigidbody classe, there is no possible way of achieving what you want. You will need to attach a script to it, in order to test for colisions :)

Hope this helps,
Benproductions1

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avatar image el_kloklo · Jun 07, 2013 at 06:46 AM 0
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Okay, I will try by using my own class so.

Thanks for the help by the way.

avatar image Benproductions1 · Jun 12, 2013 at 05:29 AM 0
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Please accept my answer then

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