Show box if radio button checked using jQuery

问题

I have the following code:

    <fieldset>
        <legend>Do you currently have SolidWorks</legend>

        <ul>
            <li><label for=""><input type="radio" name="solidworks" value="Yes" id="rdYes" /> Yes</label></li>
            <li><label for=""><input type="radio" name="solidworks" value="No" id="rdNo" /> No</label></li>
        </ul>
    </fieldset>

    <fieldset id="boxReseller" style="display:none;">
        <legend>Who is your SolidWorks reseller?</legend>
        <ul>
            <li><label for=""><input type="radio" name="reseller" value="Cad Connect" /> Cad Connect</label></li>
            <li><label for=""><input type="radio" name="reseller" value="Cadtek" /> Cadtek</label></li>
            <li><label for=""><input type="radio" name="reseller" value="CCSL" /> CCSL</label></li>
            <li><label for=""><input type="radio" name="reseller" value="Innova" /> Innova</label></li>
            <li><label for=""><input type="radio" name="reseller" value="NT CAD/CAM" /> NT CAD/CAM</label></li>
            <li><label for=""><input type="radio" name="reseller" value="Solid Engineer" /> Solid Engineer</label></li>
            <li><label for=""><input type="radio" name="reseller" value="Solid Solutions Ireland" /> Solid Solutions Ireland</label></li>
            <li><label for=""><input type="radio" name="reseller" value="Solid Solutions Management" /> Solid Solutions Management</label></li>
            <li><label for=""><input type="radio" name="reseller" value="TMS Scotland" /> TMS Scotland</label></li>
        </ul>

    </fieldset>

What I want to do is hide the second fieldset by default and if a person chooses Yes then the box will appear, and if they choose No or Yes is not selected then the box will hide again.

Can anyone help? Thanks.

回答1:

Ref Nick Craver - Nice solution although it is more felxaible as below

  $("input[name='solidworks']").change(function() {
    $("#boxReseller").toggle();
 });​​​​​​
$("input[name='solidworks']:checked").change(); //trigger correct state onload

By leaving the toggle as a wild card (undefined whichever terminology you prefer) I found it worked more efficiently. Nice though, thanks :)

回答2:

You could do this:

$("input[name='solidworks']").change(function() {
  $("#boxReseller").toggle(this.value == "Yes");
});​​​​​​
$("input[name='solidworks']:checked").change(); //trigger correct state onload

You can give it a try with the markup in the question here, and try the pre-checked "Yes" version here.

回答3:

Demo

http://jsfiddle.net/Wpt3Y/

jQuery(function(){
        jQuery("input[name=solidworks]").change(function(){          


            if ($(this).val() == "Yes") {
            jQuery("#boxReseller").slideDown()
            }
            else {
            jQuery("#boxReseller").slideUp();
            }                                                            
       });
});

​

回答4:

$(document).ready(function() {
    $("input[name=solidworks]").change(function() {
        if ($this).val() == "Yes") {
            $("#boxReseller").slideDown('fast');
        }
        else {
            $("#boxReseller").hide('fast');
        }
    })
})

回答5:

Somthing like (Sorry if there are typos, my coffee isn't made yet).:

 <fieldset id="fs">
        <legend>Do you currently have SolidWorks</legend>

        <ul>
            <li><label for=""><input type="radio" name="solidworks" value="Yes" id="rdYes" /> Yes</label></li>
            <li><label for=""><input type="radio" name="solidworks" value="No" id="rdNo" /> No</label></li>
        </ul>
    </fieldset>

There are ways of making this look better, but my coffee is not made. will edit once I get my elixer.

<script>attr('checked','checked')
     $("#fs:checkbox").click(function(){
              if($("#rdYes:checked").attr('checked','checked'))
              {
                 $("#boxReseller").css('display', 'block'); })
              }          
     });
</script>

来源：https://stackoverflow.com/questions/3791038/show-box-if-radio-button-checked-using-jquery