jQuery decimal validation with random value not correct

Question

How to validate decimal number with random value from input box using jQuery?

I have sample code below:

$("#actualWeight").on('keyup', function(e) {
  if (e.keyCode == 13) {
    var actualWeight = $("#actualWeight").val();
    var maxGrossWeight = "3.6";
    var minGrossWeight = "2.4";

    if (actualWeight > maxGrossWeight) {
      alert("More than max gross");
    } else if (actualWeight < minGrossWeight) {
      alert("Lower than min gross");
    } else {
      alert("Success");
    }
  }
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
<input type="text" id="actualWeight" />

As you can see:
Min Gross = 2.4
Max Gross = 3.6

When I try to input 200, it show me alert Success that should be show me alert More than max gross.

You can see it on this fiddle

When you read value from DOM, its read in string format. Try converting to numeric value. Updated Fiddle — Rajesh Dixit
– Rajesh Dixit, Commented Jan 22, 2018 at 7:30
You're comparing string's, not numbers! You must to parsefloat value and set the var's maxGrossWeight and minGrossWeght without " — Marc
– Marc, Commented Jan 22, 2018 at 7:32
Possible duplicate of Using Javascript to compare two input numbers in HTML? — Rajesh Dixit
– Rajesh Dixit, Commented Jan 22, 2018 at 7:33

Milind Anantwar · Accepted Answer · 2018-01-22 07:41:22Z

1

.val() is returning the value of input as string. Hence you are comparing string values.

You should set maxGrossWeight and minGrossWeght as float variable and parse the value of input element to float before doing the comparison :

 var actualWeight = parseFloat($("#actualWeight").val());
 var maxGrossWeight = 3.6;
 var minGrossWeight = 2.4;

Working Demo

edited Jan 22, 2018 at 7:41

answered Jan 22, 2018 at 7:29

Milind Anantwar

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