regex to validate three digit with one decimal

Question

I have written a regex to validate a number containing three digits and one decimal but it always return false in all cases.

re = new RegExp("^\d{1,3}\.*\d{0,1}$");
re.test(33.0);

I tried many combinations but its not working. Aim is to validate a number at max 3 digit with one decimal and less than 100. I can handle the less than 100 part.

Answer 1

You can use below code::

re = new RegExp("^[0-9]\d{0,2}(?:\.\d{0,1})?$");
re.test(33.0);

REGEX DEMO

Answer 2

This one does what you want:

 var test = [ 123, 12.34, 100.1, 100, 12, 1.2, 1 ]; console.log(test.map(function (a) { return a+' :'+/^(?:100(?:\\.0)?|\\d{1,2}(?:\\.\\d)?)$/.test(a); })); 

Answer 3

Better use regex literal:

var re = /^\d{1,2}(\.\d)?$/;
re.test(33.0);

When passing in numbers beware of floating point precision problems:

re.test(0.1 * 302); // false, because 0.1 * 302 results in 30.200000000000003 :-(

Answer 4

Remember to escape the \\ you regex sequence :

re = new RegExp("^\\\\d{1,3}\\.*\\\\d{0,1}$");

re.test(33.0);

Answer 5

Instead of using Regex Expression Object try using direct javascript

/\d{1,3}\.*\d{0,1}$/.test(33.0)

 console.log(/\\d{1,3}\\.*\\d{0,1}$/.test(33.0)) 

If you want to use RegExp

then try

new RegExp(/\d{1,3}\.*\d{0,1}$/,'i');
re.test(33.0);

 let re = new RegExp(/\\d{1,3}\\.*\\d{0,1}$/,'i'); console.log(re.test(33.0));