Question

我正在尝试创建一个小网站，从提示中询问用户号码。然后会在div标签上显示一条消息，告诉用户他们获得了什么等级。它不起作用。

例如：如果我得到90％，会出现一条消息，告诉我我的成绩是A +。

以下是代码：

<h1>Prelab1 Ex2</h1><br>
<div id="div2"></div>
<script>

  var inputperc = prompt("Enter a percentage");


  if(inputperc > 100 || inputperc < 0 || isNaN(inputperc)){
    document.getElementById("div2").innerHTML = "Enter a valid Percentage value  between 0 and 100";
    else{
    if(inputperc >= 90 && inputperc <= 100){
        document.getElementById("div2").innerHTML = "A+";
        }
        if(inputperc>90 && inputperc<100){
        document.getElementById("div2").innerHTML = "A+";
      }
      if(inputperc>85 && inputperc<89){
        document.getElementById("div2").innerHTML = "A";
      }
      if(inputperc>80 && inputperc<84){
        document.getElementById("div2").innerHTML = "A-";
      }
      if(inputperc>77 && inputperc<79){
        document.getElementById("div2").innerHTML = "B+";
      }
      if(inputperc>73 && inputperc<76){
        document.getElementById("div2").innerHTML = "B";
      }
      if(inputperc>70 && inputperc<72){
        document.getElementById("div2").innerHTML = "B-";
      }
      if(inputperc>67 && inputperc<69){
        document.getElementById("div2").innerHTML = "C+";
      }
      if(inputperc>63 && inputperc<66){
        document.getElementById("div2").innerHTML = "C";
      }
      if(inputperc>50 && inputperc<62){
        document.getElementById("div2").innerHTML = "C-";
      }
      if(inputperc>57 && inputperc<59){
        document.getElementById("div2").innerHTML = "D+";
      }
      if(inputperc>53 && inputperc<56){
        document.getElementById("div2").innerHTML = "D";
      }
      if(inputperc>50 v&& inputperc<52){
        document.getElementById("div2").innerHTML = "D-";
      }
      if(inputperc>0 && inputperc<49){
        document.getElementById("div2").innerHTML = "f";
      }
      }
    }

</script>

Answer 1

每当用户在输入字段中输入文本时，它都被视为string。所以你需要将它转换为integer。提示周围的parseInt是最简单的方法。 10只是说你正在使用0,1,2,3,4,5,6,7,8,9（base10）......不像0100100101（base2）。

var inputperc = parseInt(prompt("Enter a percentage"), 10);

Answer 2

小错误：

 ...
  if(inputperc > 100 || inputperc < 0 || isNaN(inputperc)){
        document.getElementById("div2").innerHTML = "Enter a valid Percentage value  between 0 and 100";
    } else { // <-- forgot to close the if.
    if(inputperc >= 90 && inputperc <= 100){
 ...

同样如前所述，将inputperc转换为带parseInt()的数字是个好主意。另外，请确保最后删除额外的}。

Answer 3

您的代码有一个语法错误，正如spencer指出的那样。您应该考虑编写getGrade函数。

var getGrade = function (percent) {
    if (percent > 100 || percent < 0 || isNaN(percent)) {
        return "Enter a valid Percentage value  between 0 and 100";
    }
    if (percent >= 90 && percent <= 100)    return "A+";
    if (percent > 90 && percent < 100)      return "A+"; 
    if (percent > 85 && percent < 89)       return "A";
    if (percent > 80 && percent < 84)       return "A-"; 
    if (percent > 77 && percent < 79)       return "B+";
    if (percent > 73 && percent < 76)       return "B";
    if (percent > 70 && percent < 72)       return "B-";
    if (percent > 67 && percent < 69)       return "C+";
    if (percent > 63 && percent < 66)       return "C";
    if (percent > 50 && percent < 62)       return "C-";
    if (percent > 57 && percent < 59)       return "D+";
    if (percent > 53 && percent < 56)       return "D";
    if (percent > 50 && percent < 52)       return "D-";
    if (percent > 0 && percent < 49)        return "F";
    return "Unknown Error"
};

var el      = document.getElementById("div2"),
    percent = prompt("Enter a percentage:"),
    grade   = getGrade(percent);

el.innerHTML = grade;

我的代码无效

3 个答案: