Question

我试图获得一个计数器来记录获得双倍所需的最少数量的掷骰，然后当他们玩＃39;游戏时再次，如果他们的双打数量更少，则用新的替换第一个答案。

 count=0, doubles=0, lowest=10000000;
  function Roll()
 // Assumes: die images are in http://dave-reed.com/book/Images
 // Results: displays a randomly selected image of a 6-sided die
 {
   var roll1, roll2;

   roll1 = RandomInt(1, 6);
   roll2 = RandomInt(1, 6);
   document.getElementById('die1').src = 
         "http://dave-reed.com/book/Images/die" + roll1 + ".gif";
   document.getElementById('die2').src = 
         "http://dave-reed.com/book/Images/die" + roll2 + ".gif";
   count += 1;
   if (roll1==roll2) {
       doubles += 1;
   }
   if (doubles==3) {
       alert("You got three pairs of doubles!!! It only took you " +count+ " rolls to get it! Now, go directly to jail, do not pass go, and do not collect $200");
       doubles=0;
       count=0;

   }
   if (doubles==3 && lowest>count){
           lowest=count
           document.getElementById('lowestLine').innerHTML = count;
   }
   document.getElementById('countLine').innerHTML = count;
   document.getElementById('doublesLine').innerHTML = doubles;

 }

Answer 1

一旦它变为3，你就将operator设置为0，然后检查doubles是否等于3，这是永远不会的。

看起来你应该把

doubles

在第一次if (lowest > count) { lowest = count document.getElementById('lowestLine').innerHTML = count; }内检查，但在将全局变量设置为doubles === 3之前。

像这样：

javascript计数器不显示最少数量的双打

1 个答案: