Hello Unity3D i have a question about using the same script for more than one character?Is it possible to use the same script for more than one player?the reason why i ask this question is the fact that i have more than one character that needs to do combos and so far i only got one character to do so but i want another one to do it as well.If anyone know i can fix this problem.Can you please tell me how can i?
Script if anyone was wondering
#pragma strict
var combos : combosequence[] = new combosequence[3]; //Combo class. Fill this out in Inspector.
var comboMaxTime : float = 1.0; //How long a timer for combo lasts in seconds. Fill this out in Inspector.
var comboSpanTime : float = .25; //The span of when a combo can start. Fill this out in Inspector.
var comboMidPoint : float = .5; //The position within the maxtime of a combo the span is active. Fill this out in Inspector.
private var currentCombo : int = -1;
private var comboTimeout : float = .0;
AudioListener.volume = 100;
function Update () {
if (Input.GetKeyDown("u")) Attack();
if (comboTimeout > 0) DecreaseTime();
}
function Attack () {
ComboTime(); //Do all the logic for timing before animation etc.
animation.Play(combos[currentCombo].comboAnimation); //Play the currentCombo combos' animation
audio.clip = combos[currentCombo].comboSound; //Set the audio clip to the currentCombo combos' audio
audio.Play(); //Finally play the audio
audio.volume = 50;
//Add logic for hitting enemies etc here!
}
function ComboTime () {
if (currentCombo<combos.Length-1 &&
comboTimeout>0 &&
comboTimeout>comboMidPoint-comboSpanTime &&
comboTimeout<comboMidPoint+comboSpanTime ||
currentCombo==-1) {
currentCombo++;
} else {
currentCombo = -1;
}
comboTimeout = comboMaxTime;
}
function DecreaseTime () {
comboTimeout-=1*Time.deltaTime;
}
class ComboSequence {
var comboName : String; //A name for current attack (might be of use later for tutorial or something else)
var comboAnimation : String; //The animation name for current attack
var comboSound : AudioClip; //The audio for current attack
}