CODEWITHSUNDEEP
javascripthtml
Philippe Katshianga
Philippe Katshianga

Reputation: 9

my codes aren't working

I am trying to creat a little website that ask user a number from a prompt. then a message will appear on a div tag that tell users what grade they got. it doesn't work.

Exemple: if I got 90%, a message will appear and tell me that my grade is A+.

here are the codes:

<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>

Upvotes: -1

Views: 79

Answers (3)

Ilia Choly
Ilia Choly

Reputation: 18557

Your code has a syntax error as spencer pointed out. You should consider writing a getGrade function though.

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;

Upvotes: 0

Spencer Wieczorek
Spencer Wieczorek

Reputation: 21565

Small error:

 ...
  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){
 ...

Also as already mentioned it's a good idea to convert inputperc to an Number with parseInt(). Also make sure to remove the extra } at the end because of this.

Upvotes: 0

CRABOLO
CRABOLO

Reputation: 8793

Whenever a user enters text into an input field it is considered a string. So you need to convert it to an integer. parseInt around your prompt is the easiest way. the 10 is just saying that you're using 0,1,2,3,4,5,6,7,8,9 (base10)... not something like 0100100101 (base2).

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

Upvotes: 1

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