Javascript: compare string numbes above 10.0

Question

After ajax query I have the next numbers:

var a = "5.20";
var b = "7.20";
var c = "11.20";

I have noticed that if compare numbers above 10.00 the condition is false. I do not understand why this happens

if (b > a) console.log("higher"); // true
if (c > b) console.log("higher"); // not true

I have solved the problem with next code (it should also work parseFloat)

if ((+c) > (+b)) console.log("higher"); // true

It happens because you are comparing strings which is based on lexicographical order in which "7" > "1" — Alex K., Mar 08 '18 at 12:05

score 3 · Answer 1 · answered Mar 08 '18 at 12:05

3

"5" > "10" for the same reason that "m" > "ba". Strings are compared alphabetically (even if they contain digits) - or rather lexicographically.

answered Mar 08 '18 at 12:05

melpomene

84,125
8
85
148

very good explanation – kurtko Mar 08 '18 at 12:06

score 0 · Answer 2 · answered Mar 08 '18 at 12:13

0

The variables are string type. Not integers.

typeof("5.20"); // OUTPUT: "string"

Remove the quotation marks:

var a = 5.20;
var b = 7.20;
var c = 11.20;
if (b > a) console.log("higher"); // true
if (c > b) console.log("higher"); // also true

Example: https://www.w3schools.com/js/js_comparisons.asp

answered Mar 08 '18 at 12:13

Gastón Corti

51
1
6

I know, that's why in the headline I put 'string numbers' – kurtko Mar 08 '18 at 12:15

score -1 · Answer 3 · answered Mar 08 '18 at 12:06

-1

You are comparing two strings alphabetically. If you remove the quotes where you create the variables then you will have numbers and your code will work.

var a = 5.20;
var b = 7.20;
var c = 11.20;

answered Mar 08 '18 at 12:06

Bijan Rafraf

393
1
7

Javascript: compare string numbes above 10.0

3 Answers3