jQuery if condition for radio button name and value

Question

I have a radio button like

<input type="radio" name="user-type" value="Store">
<input type="radio" name="user-type" value="Brand">

I tried writing jQuery script like

if($("input[name=user-type]:checked").val()) == "Brand"){
     $(".showstore").hide();
     $(".showbrand").show();
}

and also tried

if($("input[name=user-type]:checked").val()) == "Brand").click(function(){
       $(".showstore").hide();
     $(".showbrand").show();
});

and also tried

if ( $("input[name=user-type]:radio").val() == "Brand"){
     $(".showstore").hide();
     $(".showbrand").show();
}

None of these worked, any correction is appreciated. Thanks.

Update1

I tried in this way and this hide both

if($("input[name=user-type]:checked").val() == "Brand"){
     $(".showstore").hide();
     $(".showbrand").show();
}
else if($("input[name=user-type]:checked").val() == "Store"){
     $(".showstore").show();
     $(".showbrand").hide();
}

Answer 1

Try the following code - it basically listens for click on radio name=user-type, and toggles based on which radio button was clicked.

$(function () {
    $('.showstore').hide();
    $('.showbrand').hide();

    $("input[name=user-type]:radio").click(function () {
        if ($('input[name=user-type]:checked').val() == "Brand") {
            $('.showstore').hide();
            $('.showbrand').show();

        } else if ($('input[name=user-type]:checked').val() == "Store") {
            $('.showstore').show();
            $('.showbrand').hide();

        }
    });
});

A working fiddle: http://jsfiddle.net/FpUSH/1/

Answer 2

This is short

$('input:radio').change(
function(){
    if($(this).val() == 'Store'){
        $('.showbrand').hide();
        $('.showstore').show();
    }
    else{
        $('.showstore').hide();
        $('.showbrand').show();
    }
}
);