Can I find if an input field contains a numeric value in Javascript

Question

I'm trying to validate a form field and am new to javascript. I have been using !isNaN but this only seems to detect if there is absolutely no numbers. For example, in the below example if I have 'Willy, Fred5' entered into the input field it seems to detect the strings and does not return false. What I need to do is detect if there is even one numeric value in the input field. But can't seem to find an answer that works.

function fnCheckName(strName) {
    strName.style.background = "#FFFFFF";
    var nameChk = strName.value;
    if (nameChk.indexOf(', ') < 0 || nameChk == "" || !isNaN(nameChk){
        strName.style.background = "#FBEC5D";
        return false}
        else {return true}

try regexp class in JavaScript and create a regular expression for it. — Mozak
– Mozak, Commented May 21, 2014 at 4:07
Nice comedy, falcon... that links to a question marked as a duplicate! — user1932079
– user1932079, Commented May 21, 2014 at 4:07

000 · Accepted Answer · 2014-05-21 05:01:46Z

2

/\d/.test('foo5')
outputs: true

/\d/.test('foo')
outputs: false

/\d/.test('5')
outputs: true

So given those inputs and outputs, I think your modified code would be:

function fnCheckName(strName) {
  strName.style.background = "#FFFFFF";
  var nameChk = strName.value;
  if (nameChk.indexOf(', ') < 0 || nameChk == "" || !/\d/.test(nameChk)) {
    strName.style.background = "#FBEC5D";
    return false
  }
  else {
    return true
  }
}

edited May 21, 2014 at 5:01

answered May 21, 2014 at 4:49

000

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3 Comments

user2864740 Over a year ago

I think the OP want's the inverse (really non-negated) test form for "contains a digit".. but using /\d/ (or equivalent) is suitable for the task.

000 Over a year ago

I couldn't really tell, the question is pretty unclear. If they want the inverse then they can remove the !.

Danrex Over a year ago

Thanks for that. yes needed to take out the ! but it worked thank you so much.

deadulya · Accepted Answer · 2014-05-21 05:58:35Z

2

You need to add RegExp:

function fnCheckName(strName) {
strName.style.background = "#FFFFFF";
var nameChk = strName.value;
if (/\d/.test(nameChk)){
        strName.style.background = "#FBEC5D";
        return false
    }
    else {return true}

/^[a-zA-Z]+$/ - checks for only letters (true for "fdsfjklr")

/^([^0-9]*)$/ - checks for only numbers (true for "4321")

/\d/ - checks for any string with number (true for "fdslkj45")

edited May 21, 2014 at 5:58

answered May 21, 2014 at 4:47

deadulya

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Comments

Mozak · Accepted Answer · 2014-05-21 04:56:09Z

0

Try this Danrex

function fnCheckName(strName) {
    strName.style.background = "#FFFFFF";
    var nameChk = strName.value;
    var regNew = new RegExp(/^([^0-9]*)$/);
    if (regNew.test(nameChk)) {//If true i.e. no digit is in the string

    } else {//If false i.e. digit is in the string  

    }
}

or

function fnCheckName(strName) {
        strName.style.background = "#FFFFFF";
        var nameChk = strName.value;
        var regNew = new RegExp(/^([^0-9]*)$/);
        return (regNew.test(nameChk));
    };

edited May 21, 2014 at 4:56

answered May 21, 2014 at 4:46

Mozak

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Comments

Hamid Bahmanabady · Accepted Answer · 2014-05-21 04:43:18Z

-1

If you enter even one number return false else return true

eval('/^[a-zA-Z]+$/').test('sfjd');

EDIT

function fnCheckName(strName) {
        if( eval('/^[a-zA-Z]+$/').test(strName))
        {
            // this name not contains numbers
            // change background or every things that you want
            return true;
        }
        else
        {
            // this name has number or numbers
            // change background or every things that you want
            return false;

        }
    }

edited May 21, 2014 at 4:43

answered May 21, 2014 at 4:26

Hamid Bahmanabady

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2 Comments

Danrex Over a year ago

I don't even know how to put this in or what it is?

user2864740 Over a year ago

The use of eval is "just wrong" (it will actually have equivalent semantics if removed so eval is entirely unnecessary and useless here). That aside, this will fail and return "true" for input like =^_^=.

Surendheran · Accepted Answer · 2014-05-21 05:18:40Z

-1

Try this changes,

function fnCheckName(strName) {
strName.style.background = "#FFFFFF";
var nameChk = strName.value;
var status = false;
for(var i=0;i<nameChk.length;i++){
  if(isNan(nameChk[i])==false)
    status = true;
  else
    status = false;
}
return status;
}

edited May 21, 2014 at 5:18

answered May 21, 2014 at 4:24

Surendheran

1971 silver badge18 bronze badges

3 Comments

Danrex Over a year ago

I keep getting an error - uncaught type error - undefined is not a function on the if() line.

user2864740 Over a year ago

isNaN is a global function, not a Number method .. also a break/return is missing, so even with the isNaN change this is only "true" if the last character is not-NaN.

user2864740 Over a year ago

It's still wrong because of a missing break/return - status will be reset because the loop is not terminated correctly. Try with the values "1foo" and "bar2" as both of these contain numbers, but this function will only return true in one case. (This is also a very curious use of isNaN, although I believe it holds that !isNaN(x) === /\d/.test(x) for all single-character strings x ..)

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