Question

Please have a look on the following:

$('#myRadio').change(function() {           

    if($(this).is(':checked'))  {
        $(this).parent().addClass('green');
    } else {                              
        $(this).parent().removeClass('green');
    }
});

Markup lookslike somewhat as following

<table>
  <tr>
    <td>Some text 1 </td>
    <td><input type="radio" value="txt1" name="myRadio" id="text1" /></td>
    <td>Some text 2 </td>
    <td><input type="radio" value="txt2" name="myRadio" id="text2" /></td>
    <td>Some text 3 </td>
    <td><input type="radio" value="txt3" name="myRadio" id="text2" /></td>
  </tr>
</table>

When I switch radio, above javascript code applies 'green' to TD tag, which is fine. But if I change the selection to another it adds the green to another but doesnt remove the green from the previously selected radio.

How do I make it work so that selecting a radio option changes its parent TD's class to green and selecting another will reset all but add green to only the newly selected!

Can it also be made to change class of its first previous TD which contains "some text 3" etc??

Thanks.

Was it helpful?

Solution

Try something like this:

if($(this).is(':checked'))  {
    $(this).parent().parent().parent().find('.green').removeClass('green');
    $(this).parent().addClass('green');
}

This will find the table element of your current radio button grouping, find any elements with the green class, and remove the class.

Alternatively, if you only have one radio button group on the page, it would be simpler to just do:

$('.green').removeClass('green');

OTHER TIPS

You shouldn't use the change() event on radio buttons and checkboxes. It behaves a little dodgy and inconsistent across browsers (it causes problems in all versions of IE)

Use the click() event instead (don't worry about accessibility, because the click event will also be fired if you activate/select a radio button with the keyboard)

And as the others here pointed out, resetting the green is easy as well:

So simply change your code to

$('#myRadio').click(function() {           
    $(this).parents("tr").find(".green").removeClass("green");

    if($(this).is(':checked'))  {
        $(this).parent().addClass('green');
    }
});

EDIT: as requested in comment, also change the previous td:

$('#myRadio').click(function() {           
    $(this).parents("tr").find(".green").removeClass("green");

    if($(this).is(':checked'))  {
        $(this).parent().prev().andSelf().addClass('green');
    }
});

or even better, turning all td elements of the parent row green:

$('#myRadio').click(function() {           
    $(this).parents("tr").find(".green").removeClass("green");

    if($(this).is(':checked'))  {
        $(this).parents("tr").find("td").addClass('green');
    }
});

Try this:

if($(this).is(':checked')) {
    $(this).parent().siblings('td.green').removeClass('green');
    $(this).parent().addClass('green');
}

I just wrote a solution that does not care how deeply nested the radio buttons are, it just uses the name attribute of the radio buttons. This means this code would work without modification anywhere, I think - I had to write it since I have a strange situation where one radio button is nested differently than its cohorts with the same name.

$('input[type="radio"]').change( function() {
    // grab all the radio buttons with the same name as the one just changed
    var this_radio_set = $('input[name="'+$(this).attr("name")+'"]');

    // iterate through each  
    // if checked, set its parent label's class to "selected"
    // if not checked, remove the "selected" class from the parent label
    // my HTML markup for each button is  <label><input type="radio" /> Label Text</label>
    // so that this works anywhere even without a unique ID applied to the button and label
    for (var i=0; i < this_radio_set.length;i++) {
        if ( $(this_radio_set[i]).is(':checked') ) $(this_radio_set[i]).parents('label').addClass('selected');
        else $(this_radio_set[i]).parents('label').removeClass('selected');
    }
});

my proposal is :

$('#myRadio').change(function() {           
    _parent = $(this).parent();
    if($(this).is(':checked'))  {
       _parent.addClass('green');
    } else {                              
       _parent.removeClass('green');
    }
});

Here is the solution which worked great for me:

var $boxes=$('input:radio');
var $parents = $boxes.parent();
$boxes.click(function(){
    $parents.filter('.selected').removeClass('selected');
    $(this).parent().addClass('selected');
});
$boxes.filter(':checked').parent().addClass('selected');

Features:

  • Faster. Only selects list of checkboxes once.
  • Does not rely on :checked. I'm not sure if there is a certain order in which "clicked" and "changed" will be executed across browsers.
  • uses :radio instead of [type="radio"] which is jQuery recommended
  • sets class on the parent for the default value of radio button.

Hope this helps!

i think this is the best and more flexible way to do this:

forget the tables (google know many reason for that) and use wrappers like below

<div id="radio_wrapper">
    <span class="radio">
        <label for="myRadio1" class="myRadio">Some text 1 </label>
        <input type="radio" value="txt1" name="myRadio" id="myRadio1" class="myRadio" />
    </span>
    <span class="radio">
        <label for="myRadio2" class="myRadio">Some text 2 </label>
        <input type="radio" value="txt2" name="myRadio" id="myRadio2" class="myRadio" />
    </span>
    <span class="radio">
        <label for="myRadio3" class="myRadio">Some text 3 </label>
        <input type="radio" value="txt3" name="myRadio" id="myRadio3" class="myRadio" />
    </span>
</div

format with CSS like this

span.radio {
    background-color: #cccccc;
}
span.currentGreen {
    background-color: #ccffcc;
}

and using this script you can do exactly the same as you want

$('.myRadio').change(function() {           
    $('.currentGreen').removeClass('currentGreen'); // remove from all
    if ($(this).is(':checked')) {
        $(this).parent().addClass('currentGreen'); // add to current
    }
});

i don't like to use filter() and find() because they use unnecessary resources in this case.

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