如何根据复选框的值在侧边栏菜单上隐藏和显示特定链接？

Question

Is it possible to hide and show a link/icon on a sidebar menu based on checkbox value?

I have a checkbox that once enabled (value = on), the icon of that particular link on sidebar menu should be shown and once disabled will be hidden.

Hope someone could help me. Thanks!

file.html

{% elif field.type == 'checkbox' %} 
<div class="form-group"> 
<label>{{ field.prompt }}</label> 
<br> 
<input type="{{ field.type }}" class="form-control {{ field.name }}" name="{{ field.name }}"> 
</div>


$form_admin_config.validate({ 
ignore: 'input[type=hidden], .ignore', 
onkeyup: false, submitHandler: 
function(form) { 
$form_admin_config.find('input[type="checkbox"]').each( function () {
         var checkbox_this = $(this);

         if( checkbox_this.is(":checked") == true ) {
             checkbox_this.attr('value', 'on');    
         } else {
             checkbox_this.prop('checked',true);
             checkbox_this.attr('value', '');

             }
             })
             ajax_post_form($form_admin_config, 'Save Successful', function(){
             },
             function() {});
             }

        });

base.html

<li class="{% if request.mod == 'subscription_products' %}inactive{% endif %} main_menu" data-toggle="popover" data-content="Subscription Products">
<a href="{% url 'admin_app:subscription_products_index' %}" ><i class="fa fa-check-square-o"></i> <span class="nav-label"> Subscription Products </span></a></li>
{% endif %}

file.py

def admin_config_process(request):
    for key, val in request.POST.iteritems():
        if key != 'csrfmiddlewaretoken':
            admin_config = AdminConfig.getAdminConfigFromName(key)
            if admin_config is not None:
                admin_config.svalue = val
                admin_config.save()
            else:
                new_id = make_uuid()
                admin_config = AdminConfig.getAdminConfig(new_id)
                if radmin_config is not None:
                    new_id = make_uuid()
                AdminConfig(new_id, None, key, val).save()

    return jsonify({'status': 'ok'})

Answer 1

Try something like this:

 $('#chk').click(function(){ if( $(this).is(':checked') ) { $('#link1').show(); } else { $('#link1').hide(); } }); 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <input type="checkbox" value="1" id="chk" />Check <ul> <li id="link1">Link 1</li> </ul> 

You can get the checkbox value like:

$(this).val(); but this will give the same value doesn't matter whether the checkbox is checked or not. So you have to put the logic to get the value of checkbox only when it is checked.