Question

我用JavaScript制作了一个游戏，其中玩家只能猜测1到10的随机数三次，每次程序读取错误的答案时，它将显示应该是正确的答案，并告诉玩家再试一次，如果玩家得到正确答案，游戏会显示一条消息，提示您您理解正确。我设法使程序正常工作，显然，一切似乎都很好，除了一件事，即使玩家得到正确的答案，程序也不会停止，它会读取所有语句，直到结束。 如何在正确答案后停止？

var random = Math.floor(Math.random() * 10);

var random1 = Math.floor(Math.random() * 10);

var random2 = Math.floor(Math.random() * 10);

var answer = window.prompt("Make a guess from 1 to 10, you have 3 chances.");
if (answer == random) {
  alert("HORAAYYYY YOU GOT IT RIGHT!!!");
} else {
  alert("Sorry, the correct answer was " + random);
  window.prompt("Make a guess from 1 to 10, you have 2 chances left.");
}

if (answer == random1) {
  alert("HORAAYYYY YOU GOT IT RIGHT!!!");
} else {
  alert("Sorry, the correct answer was " + random1);
  window.prompt("Make a guess from 1 to 10, you have 1 chance left.");
}

if (answer == random2) {
  alert("HORAAYYYY YOU GOT IT RIGHT!!!");
} else {
  alert("Sorry, the correct answer was " + random2);
  alert("You've lost");
}

Answer 1

首先要说明的是虫子

您的代码中有错误。如果您在第一轮获胜，它将成功检测到这一点。在第二轮和第三轮中，任何胜利都是纯粹的巧合。

原因是因为您没有在每次调用时都将返回值从window.prompt分配给answer。因此answer的值永远不会从一个回合更改为另一个回合。

要解决此问题，您应该替换

window.prompt(...)

具有：

answer = window.prompt(...)

“错误”修复程序：

让我们从一个非常简单（但非常不好）的解决方案开始，并将其用作教授更好的架构设计的跳板。

您当前的代码大致如下：

if (win) {
   // say you won
} else {
   // say you lost
}



if (win) {
   // say you won
} else {
   // say you lost
}



if (win) {
   // say you won
} else {
   // say you lost
}

在清理完所有多余的内容之后，您可以清楚地看到它为什么要经历所有三个迭代：三个if / else块完全无关，彼此一无所知。每次它依次运行一个if / else块，然后运行另一个，然后运行另一个。

最简单的解决方法是确保以后的阻止仅在您丢失时运行。这很容易做到，因为我们已经知道您是否输了-当您不赢时就会发生！

if (win) {
   // say you won
} else {
   // say you lost
   if (win) {
      // say you won
   } else {
      // say you lost
      if (win) {
         // say you won
      } else {
         // say you lost
      }
   }
}

或者，使用您的random，random1和random2变量：

var random = Math.floor(Math.random() * 10);
var random1 = Math.floor(Math.random() * 10);
var random2 = Math.floor(Math.random() * 10);

var answer = window.prompt("Make a guess from 1 to 10, you have 3 chances."); 
if (answer == random) {
    alert("HORAAYYYY YOU GOT IT RIGHT!!!"); 
}
else {
    alert("Sorry, the correct answer was " + random);
    answer = window.prompt("Make a guess from 1 to 10, you have 2 chances left.");
    if (answer == random1) {
        alert("HORAAYYYY YOU GOT IT RIGHT!!!"); 
    }
    else {
        alert("Sorry, the correct answer was " + random1);
        answer = window.prompt("Make a guess from 1 to 10, you have 1 chance left.");
        if (answer == random2) {
            alert("HORAAYYYY YOU GOT IT RIGHT!!!"); 
        }
        else {
            alert("Sorry, the correct answer was " + random2);
            alert("You've lost");
        }
    }
}

这很丑，但是可以用。

跳板进入更好的设计

如您所想，向该游戏添加4或5或6回合将非常繁琐。您必须输入更多个if / else块，创建更多个随机变量，然后输入更多个警报语句。更令人讨厌的是，所有这些警报语句都包含相同的文本！

软件设计中有一个名为DRY的概念（请不要重复自己）。这意味着，如果您有两行相同的代码，则可以重写它以消除重复。

对于您而言，我们可以使用while循环来执行此操作，以检查我们是否赢得了比赛：

var youHaveWon = false;

while( ! youHaveWon ) {
    var random = Math.floor(Math.random() * 10);
    var answer = window.prompt("Make a guess from 1 to 10");
    if (answer == random) {
        alert("HORAAYYYY YOU GOT IT RIGHT!!!"); 
        youHaveWon = true;
    }
    else {
        alert("Sorry, the correct answer was " + random);
    }
}

这将使您不断猜测，直到正确为止，并且不重复任何代码。尽管这没有将您限制为只有3个猜测。为此，我们应该再引入一个变量：

var youHaveWon = false;
var guessesRemaining = 3;

while( ! youHaveWon && guessesRemaining > 0 ) {
    var random = Math.floor(Math.random() * 10);
    var answer = window.prompt("Make a guess from 1 to 10... you have " + guessesRemaining + " more guesses");
    if (answer == random) {
        alert("HORAAYYYY YOU GOT IT RIGHT!!!"); 
        youHaveWon = true;
    }
    else {
        alert("Sorry, the correct answer was " + random);
    }
    guessesRemaining = guessesRemaining - 1;
}

if ( ! youHaveWon ) {
    alert("You lost");
}

Answer 2

您可以将其包装到函数中并以正确的值返回。返回后，JS将停止执行，控件将向前移动

尝试以下

function findNumber() {

  var random = Math.floor(Math.random() * 10);

  var random1 = Math.floor(Math.random() * 10);

  var random2 = Math.floor(Math.random() * 10);

  var answer = window.prompt("Make a guess from 1 to 10, you have 3 chances.");
  if (answer === random) {
    alert("HORAAYYYY YOU GOT IT RIGHT!!!");
    return true;
  } else {
    alert("Sorry, the correct answer was " + random);
    window.prompt("Make a guess from 1 to 10, you have 2 chances left.");
  }

  if (answer == random1) {
    alert("HORAAYYYY YOU GOT IT RIGHT!!!");
    return true;
  } else {
    alert("Sorry, the correct answer was " + random1);
    window.prompt("Make a guess from 1 to 10, you have 1 chance left.");
  }

  if (answer == random2) {
    alert("HORAAYYYY YOU GOT IT RIGHT!!!");
    return true;
  } else {
    alert("Sorry, the correct answer was " + random2);
    alert("You've lost");
  }
}

Answer 3

问题是您拥有三个不同的if块，即使用户可能已经猜出了正确的数字，它们也会全部执行。我建议设置一个随机数和一个全局计数器，以跟踪剩余机会。

function validate() {
  if (answer == random) {
    alert("HORAAYYYY YOU GOT IT RIGHT!!!");
  } else {
    if (chances - 1 > 0) {
      chances--;
      answer = window.prompt("Make a guess from 1 to 10, you have " + chances + " chances.");
      validate();
    }
  }
}

var random = Math.floor(Math.random() * 10);
var chances = 3;
var answer = window.prompt("Make a guess from 1 to 10, you have " + chances + " chances.");
validate();

Answer 4

var randomNumbers = [];
var numberOfTries = 3;

for(var i=0; i<numberOfTries ; i++){
   randomNumbers.push(Math.floor(Math.random() * 10));  
}

for(var i=numberOfTries-1; i > -1 ; i--){
   var answer = window.prompt("Make a guess from 1 to 10, you have "+ parseInt(i + 1) +" chances left.");
   if(answer === randomNumbers[i]) {
      alert("HORAAYYYY YOU GOT IT RIGHT!!!");
      return; 
   }
   else {
      alert("Sorry, the correct answer was " + randomNumbers[i]);
   }  
}

正确答案后如何结束程序？

4 个答案:

首先要说明的是虫子

“错误”修复程序：

跳板进入更好的设计