Javascript if statement issue

Question

I'm having an issue with this basic conditional statement..

When the var equals 9 then the script works but once hits 10 and above it refers to "running low" which is only for the number 5 and below...please help..

Thank you.

<script type="text/javascript">


var inventory = "9";


if( inventory == "Sold Out" ){
   document.write("<b>Sold Out</b>");


}else if( inventory >= "6" ){
   document.write("<b>more than 5</b>");

}else if( inventory <= "5" ){
   document.write("<b>running low</b>");

}else{
  document.write("<b>error</b>");
}

</script>

You're comparing strings instead of numbers. The comparison isn't done the same way. — I Hate Lazy
– I Hate Lazy, Commented Oct 31, 2012 at 19:09
I removed the quotes from 6 and 5 but kept it on the 9 because it can be a string and not always an integer. — raffi keklikian
– raffi keklikian, Commented Oct 31, 2012 at 19:13
hint: "9" < "10" will always be false because "1" comes before "9" when compared alphabetically. related — jbabey
– jbabey, Commented Oct 31, 2012 at 19:48

Mr. Alien · Accepted Answer · 2012-10-31 19:09:35Z

2

You should compare numbers instead of string, and hence your condition gets wrong, try this

<script type="text/javascript">
var inventory = 9;
  if( inventory == "Sold Out" ){
     document.write("<b>Sold Out</b>");
  }
  else if( inventory >= 6 ){
   document.write("<b>more than 5</b>");
  }
  else if( inventory <= 5 ){
   document.write("<b>running low</b>")
  }
  else{
    document.write("<b>error</b>");
 }
</script>

answered Oct 31, 2012 at 19:09

Mr. Alien

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2 Comments

Korvin Szanto Over a year ago

Note that the OP is passing in strings (based on the first if statement), your solution (as with mine) will also work if inventory == "10".

Joshua Dwire Over a year ago

@raffi keklikian in addition to this code, you may also want to consider checking if the value is a string (typeof(inventory)=='string') or a number (typeof(inventory)=='number').

losthorse · Accepted Answer · 2012-10-31 19:27:38Z

As others have pointed out, the problem is that when numbers are quoted they have different meaning then when unquoted.

For example:

var a = 2;

console.log(a == "2"); //returns true
console.log(a === "2"); //returns false

I recommend using 'strict' operators in JavaScript in most cases.

Documentation can be found here.

Also, you may want to consider using a switch statement, as it is easier to read and maintain.

var feedback,
    inventory = 9;

switch (true) {
    case inventory === 'Sold Out' : feedback = 'Sold Out'; break;
    case inventory >= 6           : feedback = 'more than 5'; break;
    case inventory <= 5           : feedback = 'running low'; break;
    default : feedback = 'error'; break;
}

document.write('<b>' + feedback + '</b>');

HBP · Accepted Answer · 2012-10-31 19:09:52Z

0

You should be using numbers not strings in your comparisons.

answered Oct 31, 2012 at 19:09

HBP

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Comments

Korvin Szanto · Accepted Answer · 2012-10-31 19:10:39Z

0

Your problem is the fact that you're comparing strings rather than numbers, try using inventory >= 5

Example: http://jsfiddle.net/UwGGg/

var inventory = "10";


if ( inventory == "Sold Out" ){
   document.write("<b>Sold Out</b>");


} else if ( inventory >= 6 ){
   document.write("<b>more than 5</b>");

} else if ( inventory <= 5 ){
   document.write("<b>running low</b>");

} else {
  document.write("<b>error</b>");
}

answered Oct 31, 2012 at 19:10

Korvin Szanto

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Comments

rogodeter · Accepted Answer · 2012-10-31 19:24:16Z

0

If you want to be expressly clear that your variable (inventory) can be a string OR a number (seems you will accept the value "Sold Out"), I recommend forcing it into a number for comparison... This should reduce/eliminate unexpected results (if you do unit testing) and provide some code documentation for somebody else to key them off that you have are testing a value that is KNOWN to be either string or number.

answered Oct 31, 2012 at 19:24

rogodeter

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