typeof command doesnt work

Question

Hi I'm trying to check the value of DOM with:

if ($('#ATTN').val() == '') {               
    $('#ATTN').val(0);
} else {
    iattn=$('#ATTN').val();
    alert(iattn);

    if(typeof iattn == 'number'){
        alert('oh');
    }
}

but it returns nothing. Also no error shown.

Have you opened your debugger and seen what `typeof iattn` equals exactly? — arjabbar, Jun 30 '16 at 14:43
Seems like `typeof iattn` would be a string if you got it from an input type text. — Hanlet Escaño, Jun 30 '16 at 14:43
@TarasKumpanenko Not in *all* cases (see [How can val() return Number?](http://stackoverflow.com/a/9227362/1715579)) — p.s.w.g, Jun 30 '16 at 14:45
@p.s.w.g. I`m agree, not always. Just make a suggestion $('#ATTN')- simple input[type=text] — Taras Kumpanenko, Jun 30 '16 at 14:48
We know because we're told that no alert occurs. OK, technically all we know is that it's not a number, so it could in theory be an array, but still. — Chris Lear, Jun 30 '16 at 14:49
@ChrisLear The question is unclear, we're only told it "returns nothing", which would certainly be true since there's no `return` statement. In my opinion, OP *really* needs to clarify and add sufficient context to fully answer the question—ideally with a snippet or jsfiddle that reproduces the behavior. — p.s.w.g, Jun 30 '16 at 14:54
@p.s.w.g I think that "no error shown" means that the `alert` doesn't trigger. But I agree with your general point. — Chris Lear, Jun 30 '16 at 14:57

Chris Lear · Answer 1 · 2016-06-30T15:24:02.437

1

This line

iattn=$('#ATTN').val();

Returns a string.

If you want to see whether it can be converted to an integer, then what you want is this:

iattn=parseInt($('#ATTN').val());
if (iattn) {
    alert("oh");
}

Use parseFloat rather than parseInt if you want a more general number test.

If you want to check that the result of parsing matches the input, use something like this:

iattn=$('#ATTN').val();
if (iattn == parseInt(iattn)) {
    alert("oh");
}

A simpler check, which won't actually convert anything to a number, and will allow all number formats (eg 2.3, 0x3, +23) is this:

if (!isNaN(iattn)) {
    alert("oh");
}

Credit to @p.s.w.g from a comment.

edited Jun 30 '16 at 15:24

answered Jun 30 '16 at 14:44

Chris Lear

6,057
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but chances are there the user give a text value by mistake and parseint will convert that too to an integer . that will make mess ! – Bineesh Jun 30 '16 at 14:50
1

If the user gives a text value that can't be converted to an integer, `iattn` will be `NaN` and the if condition will fail. – Chris Lear Jun 30 '16 at 14:52
I've updated with a check that the parsed value is similar to the input, to deal with cases like "2sdfsdsdf". But then if the user enters eg "2.0" the check will fail. – Chris Lear Jun 30 '16 at 14:55
I believe `if (!isNaN(iattn))` would be equivalent for all cases except the empty string (you could use `if (iattn && !isNaN(iattn))` if necessary). – p.s.w.g Jun 30 '16 at 15:07

HudsonPH · Answer 2 · 2016-06-30T15:01:44.967

0

.val() return a DOMString(string)

so typeof iattn always will be a string

if you want the typeof number you need convert to int

using parseInt(iattn) or parseFloat(iattn)

edited Jun 30 '16 at 15:01

answered Jun 30 '16 at 14:58

HudsonPH

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typeof command doesnt work

2 Answers2