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Question by nikemikey3 · Apr 04, 2012 at 12:19 AM · javascriptbce0043unexpected token

Why is there an unexpected token error?

I am working on creating a respawn script in Javascript and I am getting three "BCE0043 Unexpected token: .." errors when I try to run the program. They are on lines 7-9, all three of the variables in the Start function. Here is my code:

 #pragma strict

 var Position = {};
 
 function Start () {
 
     var Position.startX = gameObject.transform.position.x;
     var Position.startY = gameObject.transform.position.y;
     var Position.startZ = gameObject.transform.position.z;
 
 }
 
 function Update () {
 }
 
 function OnTriggerEnter (other : Collider) {

     if (other.name == "Player"){
         
         other.transform.position = Vector3(Position.startX, Position.startY, Position.startZ);
         
     }
 }

Any help would be greatly appreciated. Thanks in advance.

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Answer by Eric5h5 · Apr 04, 2012 at 12:51 AM

Not quite sure why you're trying to use a hashtable for that; just use a Vector3.

 private var startPosition : Vector3;
 
 function Start () {
     startPosition = transform.position;
 }
 
 function OnTriggerEnter (other : Collider) {
     if (other.name == "Player") {
         other.transform.position = startPosition;
     }
 }
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Answer by nikemikey3 · Apr 04, 2012 at 12:59 AM

Actually, nevermind, I completely changed my code to work in Vector3 correctly. Here it is if this helps anyone else: #pragma strict

 var playerPosition : Vector3;
 
 function Start () {
 
     playerPosition[0] = gameObject.Find("Player").transform.position.x;
     playerPosition[1] = gameObject.Find("Player").transform.position.y;
     playerPosition[2] = gameObject.Find("Player").transform.position.z;
     
 }
 
 function Update () {
 }
 
 function OnTriggerEnter (other : Collider) {
     
     if (other.name == "Player"){
         
         other.transform.position = playerPosition;
         
     }
 }
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avatar image Eric5h5 · Apr 04, 2012 at 01:47 AM 0
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Take a look at my answer; you can just assign a Vector3, you don't need to do x, y, and z separately.

avatar image Kryptos · Apr 04, 2012 at 07:05 AM 0
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Indeed, you don't have to make manually a copy of the Vector3 because it is not a reference-type but a value-type. Thus assignation always makes a copy.

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