JavaScript: Not running on page

Question

I have some javacript that I am trying to implement on an HTML page. The code is below - how can I get it to run as soon as the .html file is opened? It want it to prompt me to enter the weight as soon as i open it.

<script type="text/javascript">
    var weight = parseInt(prompt("What is your weight?"));
    if (weight > 126) {
        alert('Please enter a weight lighter than 126');
    }
    document.writeln('<br/>');
    var wArray = ["fly", "superfly", "bantam", "superbantam", "feather"];

    function recruit() {
        if (0 < weight < 112){
            document.writeln('You are in' + wArray[0] +'class!');
        }
        if (112 < weight < 115){
            alert('You are in' + wArray[1] +'class!');
        }
        if (weight > 115 && weight < 118){
            alert('You are in' + wArray[2] +'class!');
        }
        if(weight > 118 && weight < 122){
            alert('You are in' + wArray[3] +'class!');
        }
        if (weight > 122&& weight < 126){
            alert('Your weight class is' + wArray[4]);
    }
</script>

Just a note, you should add `,10` to `parseInt`. Like this: `parseInt(prompt("What is your weight?"), 10)`. — gen_Eric, Dec 05 '11 at 22:46

score 1 · Accepted Answer · edited May 23 '17 at 11:55

You have 2 problems with the script:

Your recruit() is missing closing bracket }
You are not calling recruit() anywhere.

Several other issues :) :

112 < weight < 115 is not what you want. It will return true even when weight is 1000. It will check if weight is bigger than 112 and then compare if the result is smaller than 115.

Try:

var weight = 1000;
// 112 < 1000 => true, true < 115 => true
if (112 < weight < 115) { 
   alert("112 < weight < 115");
}

You left some values in your checks. If you check for weight < 118 in one if and weight > 118 in another if then what will happen if weight is exactly 118?
Use if { } else if { } blocks instead of multiple if-s. In your recruit function even when the weight is 110 and the body of 1st if block is executed, you are still checking the weight in every other if.
Add spaces to the output strings, e.g.: 'You are in ' + wArray[0] + ' class! :).
As Rocket wrote in the comments: parseInt() should be called with 10 as a 2nd parameter (radix) to force parsing into decimal system. See this.
Bonus: Stop using document.writeln as soon as possible as it is teaching you bad practices. For example see this or this.

HERE is a version addressing these issues.

Note: The code is executed inside window.onload to make sure that the rest of the document is already loaded so he can access the div.

score 0 · Answer 2 · answered Dec 05 '11 at 22:38

<script type="text/javascript">
  var weight = parseInt(prompt("What is your weight?"));
  if (weight > 126) {
    alert('Please enter a weight lighter than 126');
  }
  document.writeln('<br/>');
  var wArray = ["fly", "superfly", "bantam", "superbantam", "feather"];


  function recruit() {
    if (0 < weight < 112){
      document.writeln('You are in' + wArray[0] +'class!');
    }
    if (112 < weight < 115){
      alert('You are in' + wArray[1] +'class!');
    }
    if (weight > 115 && weight < 118) {
      alert('You are in' + wArray[2] +'class!');
    }
    if(weight > 118 && weight < 122){
      alert('You are in' + wArray[3] +'class!');
    }
    if (weight > 122&& weight < 126){
      alert('Your weight class is' + wArray[4]);
    }
  }
  recruit();
</script>

A little explanation would be helpful, I don't think you needed to re-post the code unless you changed it. If you did, highlight what was changed. — Wesley Murch, Dec 05 '11 at 22:40

score 0 · Answer 3 · answered Dec 05 '11 at 22:39

0

It seems to be working fine: http://jsfiddle.net/maniator/XRGHt/

You left out a } at the end.

answered Dec 05 '11 at 22:39

Naftali

144,921
39
244
303

score 0 · Answer 4 · answered Dec 05 '11 at 22:41

0

You can call a javascript function when a page loads like this:

<body onload="myFunction();">

answered Dec 05 '11 at 22:41

novacara

2,207
4
24
34

JavaScript: Not running on page

4 Answers4