Checking number using isNaN function

JavaScript has one important function isNaN() to check any data is number or string. This function isNaN() is remembered as short form is Is Not a Number. This function returns true if the data is string and returns false if the data is a number. Here is one simple example to check one string. We will use one if else condition check to display a message accordingly. Here is the code.

var my_string="This is a string";
if(isNaN(my_string)){
document.write ("this is not a number ");
}else{document.write ("this is a number ");
}

We will integrate this function to one JavaScript prompt. Here we will ask the user to enter a number. The value entered by user we will check by isNan function and display message accordingly.

Demo of isNan

Here is the code.

var my_string = prompt("Please enter a number","");
document.write(my_string)
if(isNaN(my_string)){
document.write ("this is not a number ");
}else{document.write ("this is a number ");
}

Number of User Comments : 9


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Me2

07-03-2009

Cool
tcx

11-10-2009

yes, agree, this is very useful topic it helps me in project, good concept not finding in other sites
Dhaval Shingala

25-02-2010

The simplest Way is Why u check entered input is Number or not just check on Keypress Event is number or not The Function for checking Is number is function isNumberKey(B,C){if(!C){C=0}var A=(B.which)?B.which:event.keyCode;if(A>31
Rakesh Joshi

16-04-2010

I observed that isNaN has a flaw. If we enter a value with alphabets at the start, it will detect. But if it starts with a number and continues with alphabets or ends with alphabets, then isNaN will return false, meaning to say it is a number, which is not the desired output. Is there any other alternative?

01-09-2010

write a javascript code that will pop-up when an input of not a number is enter and return true when number data is enter. but without using in inbuild function such as isNan.
angelito

09-06-2012

This is okay..
Ray

30-08-2012

Use Integer.parseInt("value") instead and put that in a try catch block.
san

11-02-2014

@Ray, this is JavaScript not Java.
kamal

07-08-2014

thanx alot very useful

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