Error when checking checkbox property

Question

I want to show div if checkbox is checked. If not, I want to hide it. Here is my code, but it doesn't work.

$(".formularz_zgloszeniowy .firma input[type='checkbox']").change(function() {
    if(this.checked) {
        $(".formularz_zgloszeniowy .formularz_firma").css('display','block');
    } 
    else 
    {
      $(".formularz_zgloszeniowy .formularz_firma").css('display','none');
    }
});

I'm getting this error:

Uncaught SyntaxError: missing ) after argument list

EDIT: Here is full code in head tag:

<script type="text/javascript">
jQuery(document).ready(function($){

$(".formularz_zgloszeniowy .firma input[type='checkbox']").change(function() {
    if($(this).is(":checked")) {
        $(".formularz_zgloszeniowy .formularz_firma").css('display','block');
    } 
    else 
    {
      $(".formularz_zgloszeniowy .formularz_firma").css('display','none');
    }
});

});
</script>

That code should not throw such error. The culprit must be something else. — Ram
– Ram, Commented Nov 23, 2015 at 19:19
There's a ton of non-printable characters in the example, probably in your real code too? — Teemu
– Teemu, Commented Nov 23, 2015 at 19:20
@gkopowski Just check the fiddle I've linked. You see the red dots? They are \u200b characters breaking your code. — Teemu
– Teemu, Commented Nov 23, 2015 at 19:26

Racil Hilan · Accepted Answer · 2015-11-23 19:44:35Z

You have some invisible characters in your code that are causing the problem. Here is your code working perfectly without those characters:

$(".formularz_zgloszeniowy .firma input[type='checkbox']").change(function(){
  if(this.checked) {
    $(".formularz_zgloszeniowy .formularz_firma").css('display','block');
  } 
  else {
    $(".formularz_zgloszeniowy .formularz_firma").css('display','none');
  }
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="formularz_zgloszeniowy">
  <div class="firma">
    <input type="checkbox" value="test" /> Test checkbox
  </div>
  <div class="formularz_firma">Test div</div>
</div>

And here is the code copied from your question with those invisible characters, so it is not working:

$(".formularz_zgloszeniowy .firma input[type='checkbox']").change(function() {
    if(this.checked) {
        $(".formularz_zgloszeniowy .formularz_firma").css('display','block');
    } 
    else 
    {
      $(".formularz_zgloszeniowy .formularz_firma").css('display','none');
    }
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="formularz_zgloszeniowy">
  <div class="firma">
    <input type="checkbox" value="test" /> Test checkbox
  </div>
  <div class="formularz_firma">Test div</div>
</div>

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