检查碰撞功能匹配相同位置的2个对象

Question

完整代码here

我注意到如果在同一个位置有两个错误并且玩家发生碰撞，它会因为它在技术上与两个不同的物体发生碰撞而计算两次死亡，我该如何改变它以检查碰撞1一个时间，还是只算1死？

checkCollision(playerl,playerr) {
  //check collision for each bug
  for (var i = 0; i < 5; i++) {
        var thisEnemy = allEnemies[i];
        if (
           thisEnemy.leftLimit < player.rightLimit &&
           thisEnemy.rightLimit > player.leftLimit &&
           thisEnemy.upperLimit > player.lowerLimit &&
           thisEnemy.lowerLimit < player.upperLimit) {
           console.log("collision");
           //console.log(player.lowerLimit, player.upperLimit, thisEnemy.lowerLimit, thisEnemy.upperLimit)
          player.loseLife();
       }
    }
};

Answer 1

为什么不在死亡一次后使用setTimeout来实现短暂的无敌？

//somewhere in your code
var dying = false;

checkCollision(playerl,playerr) {
  //check collision for each bug
  for (var i = 0; i < 5; i++) {
        var thisEnemy = allEnemies[i];
        if (
            thisEnemy.leftLimit < player.rightLimit &&
            thisEnemy.rightLimit > player.leftLimit &&
            thisEnemy.upperLimit > player.lowerLimit &&
            thisEnemy.lowerLimit < player.upperLimit &&
            dying==false) {

            console.log("collision");
            dying=true; // dying is true, so we wont end up in this block again
            setTimeout(function(){
                dying=false; //after 500 milliseconds we set dying to false so our player has the ability to die again!
            },500);
            player.loseLife();
       }
    }
};