Random Teleport Location

Questions & Answers
1

I’m extremely new to scripting and need help! I’m using the standard

var destination : Transform;

function OnTriggerEnter(other : Collider) {
if (other.tag == “EvilFace”) {
other.transform.position = destination.position;
audio.Play();
}

}

“Teleport Code” (with audio!)but I have been unable to figure out how to code a random location into the teleport (or a random audio, but that’s another problem for another question). I’ve tried a random target variable but that didn’t seem to do the job. I know this is probably an extremely easy question, but I have problems wrapping my brain around scripting. If anybody can help me, and maybe explain the how and why of the random scripting? I know I’m going to be doing a lot of random in the future. Thanks!

4

Right now you are teleporting to wherever you placed the destination transform. If thats null, itll probly teleport you to Vector3.zero (Vector3(0,0,0)). To get a completely random position you can use Random.Range() to put in a min and max for the range.

destination.transform.position=Vector3(Random.Range(0,300),
                                       Random.Range(0,300),Random.Range(0,300));

this will place destination at some random X,Y,&Z position between 0 and 300 (Vector3(50,279,133) as an example). Very doubtful this will be what you want, but it should help with the whole “Random-ness” situation.

You should just make that the actual destination a Vector3 and randomize it in the trigger function.

var destination:Vector3;

function OnTriggerEnter(other : Collider) 
{ 
    if(other.tag == "EvilFace") 
    {
        destination=Vector3(Random.Range(0,300),
                                       Random.Range(0,300),Random.Range(0,300));
        other.transform.position = destination; 
        audio.Play(); 
    }
}
2

You need a math function to generate a random number for each vector position.

int num = random.Next(1000); //whatever number you pass is the max

destination.position.x=num;

num = random.Next(1000);

destination.position.y=num;

num = random.Next(1000);

destination.position.z=num;

What this does is generate a random number then assign it to X, then do it again for Y, and then Z so you get a random XYZ.

3

A position is made up of three coordinates (x, y and z) represented by a Vector3 data structure. Unity provides an API called Random that can generate a random number between two values. If we use this API to generate random new values for the x, y and z properties of the Game Object’s position, the Game Object will effectively have moved to a new position. Your challenge will be scripting correctly to choose the range values for the random numbers such that the bounds of the values are where you want them … otherwise you could teleport your character to an offscreen position and never find him again.

See:

http://unity3d.com/support/documentation/ScriptReference/Random.Range.html

5

Bingo! I’m not sure if it was changing it from Random.Range to Random.RandomRange, or if I just needed to tweak the values to fit my map size/location, but the following did the trick perfectly. It still tends to favor a certain side of the map, but not nearly as much.

var destination : Vector3;

function OnTriggerEnter(other : Collider) { 
	if (other.tag == "EvilFace") 
		{ 
		destination = Vector3(Random.RandomRange(-500,1000), Random.RandomRange(2,10),Random.RandomRange(-500,1000));
		other.transform.position = destination; 
		audio.Play(); 
		}

}