CODEWITHSUNDEEP
javascriptregex
photonacl
photonacl

Reputation: 135

How to check if a string is a positive integer, allowing 0 and all-0 decimals

So to check if a string is a positive integer, I have done some research and found this solution here by someone: Validate that a string is a positive integer

function isNormalInteger(str) {
    return /^\+?(0|[1-9]\d*)$/.test(str);
}

However, I put this into test, and found that numbers with pure 0's on the decimal places does not seem to be working. For example:

15 ===> Works!

15.0 ====> Does not work :(

15.000 ===> Does not work :(

Build upon the existing method, how could I allow pure-0's on the decimal places and make them all work? Please note 15.38 should not work, but 15.00 should.

Upvotes: 1

Views: 389

Answers (3)

Bob Stiles
Bob Stiles

Reputation: 11

Quick and dirty

function isNormalInteger(str) {
    var ival=parseInt(str);
    return ival!=NaN && ival>=0 && ival==parseFloat(str);
}

Upvotes: 1

cнŝdk
cнŝdk

Reputation: 32145

First of all the function should be called isNormalNumber instead of isNormalInteger as it accepts decimals, then this is the REgex you need:

    function isNormalNumber(str) {
      return /^\+*[0-9]\d*(\.0+)?$/.test(str);
    }

    alert(isNormalNumber("+10.0") + "::" + isNormalNumber("+10.9") + "::" + isNormalNumber("10"));

Returns true::false:true.

EDIT:

This is an edit to avoid matching leading zeros like in the numbers 001234 and 07878:

^\+*[1-9]\d*(\.0+)?$|^0(\.0+)?$

Upvotes: 1

anubhava
anubhava

Reputation: 785156

No need to use regex here.

function isNormalInteger(str) {
   var n = parseInt(str);
   return n > 0 && n == +str;
}

Then test it:

isNormalInteger(15)
true
isNormalInteger(15.00)
true
isNormalInteger(15.38)
false
isNormalInteger(-15)
false
isNormalInteger(-15.1)
false

Upvotes: 3

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