更好的方式來繼續而不是重複

Question

我認爲有一個更好的方法，然後重複每個人一遍又一遍的相同的過程？

和碰撞方法，碰撞方法只適用於中心到中心的碰撞，這是非常好的。

var ereset = function(){ 
    //Attacker 1 
    att1.x = 0 + (Math.random() * (canvas.width - 64)); 
    att1.y = -60 ; 
    //Attacker 2 
    att2.x = 0 + (Math.random() * (canvas.width - 64)); 
    att2.y = -60 ;  
    //Attacker 3 
    att3.x = 0 + (Math.random() * (canvas.width - 64)); 
    att3.y = -60 ; 
    //Attacker 4 
    att4.x = 0 + (Math.random() * (canvas.width - 64)); 
    att4.y = -60 ; 
    //Attacker 5 
    att5.x = 0 + (Math.random() * (canvas.width - 64)); 
    att5.y = -60 ; 
} 



    if (
     hero.x <= (att1.x + 20 || att1.x + 32) 
     && att1.x <= (hero.x + 20 || att1.x + 32) 
     && hero.y <= (att1.y + 20 || att1.y - 32) 
     && att1.y <= (hero.y + 20 || att1.y - 32) 
    ){ 
     end(); 
    } else if(
     hero.x <= (att2.x + 20 || att2.x + 32) 
     && att2.x <= (hero.x + 20 || att2.x + 32) 
     && hero.y <= (att2.y + 20 || att2.y - 32) 
     && att2.y <= (hero.y + 20 || att2.y - 32) 
    ){ 
     end(); 
    }else if(
     hero.x <= (att3.x + 20 || att3.x + 32) 
     && att3.x <= (hero.x + 20 || att3.x + 32) 
     && hero.y <= (att3.y + 20 || att3.y - 32) 
     && att3.y <= (hero.y + 20 || att3.y - 32) 
    ){ 
     end(); 
    }else if(
     hero.x <= (att4.x + 20 || att4.x + 32) 
     && att4.x <= (hero.x + 20 || att4.x + 32) 
     && hero.y <= (att4.y + 20 || att4.y - 32) 
     && att4.y <= (hero.y + 20 || att4.y - 32) 
    ){ 
     end(); 
    } 

}; 

Answer 1

由於您對每個攻擊者都做了同樣的事情，所以最好製作一個執行該操作的函數，並針對每個攻擊者運行該函數。在這裏，我將把所有攻擊者放在一個數組中，並使用forEach來遍歷列表。

var attackers = [ 
    att1, att2, att3, att4, att5 
]; 

function ereset(){ 
    attackers.forEach(moveAttacker); 
} 

function moveAttacker(attacker){ 
    attacker.x = 0 + (Math.random() * (canvas.width - 64)); 
    attacker.y = -60 ; 
} 

Answer 2

OOP工作：

var Attacker = function() { 
    this.x = 0; 
    this.y = 0; 

    this.init = function(ix, iy) { 
     this.x = ix; 
     this.y = iy; 
    }; 
}; 

var ereset = function(){ 

    this.attackers = 5; 
    this.swat = []; 

    for (var n = 0; n < this.attackers; n++) 
    { 
     var att = new Attacker(); 
     att.init(0 + (Math.random() * (canvas.width - 64)), -60); 
     this.swat.push(att); 
    } 
}; 

var checkHero = function (hero, attackers) 
{ 
     for (var n = 0; n < attackers.length; n++ ) 
     { 
     if (
      hero.x <= (attackers[n].x + 20 || attackers[n].x + 32) 
      && attackers[n].x <= (hero.x + 20 || attackers[n].x + 32) 
      && hero.y <= (att1.y + 20 || attackers[n].y - 32) 
      && attackers[n].y <= (hero.y + 20 || attackers[n].y - 32) 
      ){ 
       end(); 
     } 
     } 
}; 

或這樣的事情，這不是測試的代碼，只是一種方法！