Suppress JavaScript undefined errors?

Question

I have written a script that checks a set of radiobuttons to be checked. But due to different possibilities different radiobuttons will show. Is there a way to suppress JavaScript errors when it pops undefined/getElementById is null? Something like the @-char does in PHP?

Update:

A bit more background info. I've made a website where users can submit images and another party for whom the images are can select their top 3 of the images. So each image has three radiobuttons. The difficulty here lies in the fact that the radiobuttons must be controlled dimensional (horizontal and vertical), because a submitted image may only be at place 1, 2 or 3. This is my working code. But adding many if(!var == undefined) doesn't make the code prettier. Therefor I'm wondering if there is something like @suppressMe is possible?

function HandleRadioButtons(id, type, idString, img)
{
    var idArray = idString.split("|");  
    var place1  = document.getElementById("G_" + id);
    var place2  = document.getElementById("S_" + id);
    var place3  = document.getElementById("B_" + id);
    var img1    = document.getElementById("Winner1");
    var img2    = document.getElementById("Winner2");
    var img3    = document.getElementById("Winner3");    

    switch(type)
    {
        case "G" :
            place2.checked = false;
            place2.disabled = true;
            place3.checked = false;
            place3.disabled = true;
            img1.style.background = 'url(' + img + ') no-repeat center center #FFF';
            break;
        case "S" :
            place1.checked = false;
            place1.disabled = true;
            place3.checked = false;
            place3.disabled = true;
            img2.style.background = 'url(' + img + ') no-repeat center center #FFF';
            break;
        case "B" :
            place1.checked = false;
            place1.disabled = true;
            place2.checked = false;
            place2.disabled = true;
            img3.style.background = 'url(' + img + ') no-repeat center center #FFF';
            break;
    }     

    var current1, current2, current3 = "";

    for(i = 0; i < idArray.length - 1; i++)
    {
        var place1 = document.getElementById("G_" + idArray[i]);
        var place2 = document.getElementById("S_" + idArray[i]);
        var place3 = document.getElementById("B_" + idArray[i]);

        if(place1.checked == true)
        {
            var current1 = idArray[i];            
        }

        if(place2.checked == true)
        {
            var current2 = idArray[i];            
        }

        if(place3.checked == true)
        {
            var current3 = idArray[i];            
        }
    }

    for(i = 0; i < idArray.length - 1; i++)
    {        
        var place1 = document.getElementById("G_" + idArray[i]);
        var place2 = document.getElementById("S_" + idArray[i]);
        var place3 = document.getElementById("B_" + idArray[i]);

        if(idArray[i] != id && idArray[i] != current1 && idArray[i] != current2 && idArray[i] != current3)
        {
            switch(type)
            {
                case "G" :
                    place1.disabled = false;
                    place2.disabled = false;
                    place3.disabled = false;
                    break;
                case "S" :
                    place1.disabled = false;
                    place2.disabled = false;
                    place3.disabled = false;
                    break;
                case "B" :
                    place1.disabled = false;
                    place2.disabled = false;
                    place3.disabled = false;
                    break;
            }
        }
    }   
}

Why don’t you check if they are undefined or the returned value is null? — Gumbo
– Gumbo, Commented Nov 4, 2010 at 21:49
Sounds like you'd be better off fixing the root of the problem instead. What I mean is that maybe there's a way to code it so that you can avoid it entirely. Either way, you should post your code. — Radu
– Radu, Commented Nov 4, 2010 at 21:49

Daniel Vassallo · Accepted Answer · 2010-11-04 22:02:23Z

4

You can easily test for a null or undefined value in JavaScript, as both these values are falsy:

var element = document.getElementById('some-id');
if (element) {
   element.value = 'Hello';
}

You could also consider using a try/catch block:

try {
   var element = document.getElementById('some-id');
   element.value = 'Hello';

   // ... the rest of your code here.
}
catch (e) {
   if (!(e instanceof TypeError)) {
      // The exception is not a TypeError, so throw it again.
      throw e;
   }
}

However be careful that the above will suppress all the TypeError exceptions and that might make your code more difficult to debug.

edited Nov 4, 2010 at 22:02

answered Nov 4, 2010 at 21:52

Daniel Vassallo

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2 Comments

Ben Over a year ago

Okay, that's still with IF's thus.. Well, if there is no other option, that should do it ;) Thanks

Ben Over a year ago

I'll try that tomorrow, thanks! +1 from me, if it works you'll get the accept too ;) I'll check in tomorrow again.

Mohamed Fazal · Accepted Answer · 2021-01-28 03:00:43Z

1

Try this

var element = document.getElementById('some-id');
element?.value = 'Hello';

answered Jan 28, 2021 at 3:00

Mohamed Fazal

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1 Comment

blackcatweb Over a year ago

Note that optional chaining won't work in NodeJS before v 14.

Phaenotyp · Accepted Answer · 2010-11-04 21:52:06Z

0

you can check if buttons are existing by getElementById and then check its length. Are you using any framework?

answered Nov 4, 2010 at 21:52

Phaenotyp

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