Question

我正在写一个数字猜谜游戏。每次我运行文档时，它会自动转到我的外部声明，我知道原因，但我无法弄清楚如何让它等待文本框的输入值。我想设置＆＃34;猜测＆＃34; =按提交后输入文本框的值，然后输入if语句。此时它会自动设置为null并导致错误。这是代码。

<head>
        <title>Guessing Game</title>
</head>

<body>
        <h1>Number Guessing Game</h1><br>
        <button onclick = "search(1,100)">Press to play!</button>
</body>

<script type="text/javascript">

    var counter = 0;
    var randomNumber = Math.floor(Math.random()*100+1);
    function search(first, last){
                document.open();
                document.write("<h1>Number Guessing Game</h1><br>");
                document.write("<input id = 'guessBox' type = 'text'>&nbsp;");
                document.write("<input type = 'submit' id = 'submit' value = 'Submit'><br>");
                document.write("<h2>Numbers Left</h2>");
                for(var i = first; i <= last; i++){
                    document.write(i + " ");
                }
                document.write("<br><br><h3>Number of tries: " + counter + "</h3>");
                document.close();
                var guess = document.getElementById('guessBox').value;
                //var guess = prompt("Guess!");
                myguess = parseInt(guess);
        if(myguess <= last && myguess >= first && cont == true){
            if(myguess == randomNumber){
                counter++;
                if(counter <=3){
                alert("WOW, that was amazingly quick! You found it in " + counter + " tries.");
                }
                else if(counter < 6 && counter >= 4){
                    alert("Not bad! You found it in " + counter + " tries.");
                }
                else if(counter < 10 && counter >= 6){
                    alert("Ouch! You found it in " + counter + " tries.");
                }
                else{
                    alert("Having a bad day aren't you? You found it in "+ counter + " tries");
                }

            }
            else if(myguess < randomNumber){
                first = myguess+1;
                alert("Try again! The number is higher.");
                counter++;
                search(first, last);
            }
            else{
                last = myguess-1;
                alert("Try again! The number is lower.");
                counter++;
                search(first, last);        
            }
        }
        else{
            alert("Invalid Number, try again.");
            search(first, last);
        }
    }

</script>

Answer 1

您是否尝试禁用该按钮，然后在文本框中添加onChange事件，以便在获得所需输入后启用按钮？您需要在按钮中添加ID或Name值才能访问它？并添加@LGSon添加的内容：

if（myguess＆lt; = last＆amp;＆amp; myguess＆gt; = first＆amp;＆amp; cont == true）{，你检查变量cont，我在你的代码中找不到，所以如果它是未声明并设置在某处，您的代码将始终采用外部路径。

其次，您需要将搜索功能拆分为2个函数，一个生成输入，一个在有人输入值时运行（可以在按键或按钮上触发）

如何使文档等待指定的输入值？

1 个答案: