javascript - How can I make an element appear and disappear using Jquery. - Stack Overflow

I want to do something like this:

$(document).ready(function(){
    if($('.undermenu').css('display','block')){
        $('#menu').click(function(){
            $('.undermenu').css('display','none');
        });
    }
    else{
        $('#menu').click(function(){
            $('.undermenu').css('display','block');
        });
    }
});

This code does not work, but is there any Jquery "effect" or whatever that I can use to hide/un-hide. For example is there and way to check whether or not display is set to none or block? Thanks.

I want to do something like this:

$(document).ready(function(){
    if($('.undermenu').css('display','block')){
        $('#menu').click(function(){
            $('.undermenu').css('display','none');
        });
    }
    else{
        $('#menu').click(function(){
            $('.undermenu').css('display','block');
        });
    }
});

This code does not work, but is there any Jquery "effect" or whatever that I can use to hide/un-hide. For example is there and way to check whether or not display is set to none or block? Thanks.

• javascript
• jquery
• html
• css
• dom

Share Improve this question Follow edited Apr 14, 2014 at 20:03 Anubhav 7,23855 gold badges2323 silver badges3434 bronze badges asked Apr 14, 2014 at 19:59 user221146user221146 3122 silver badges55 bronze badges

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5 Answers 5

Sorted by: Reset to default 11

Just use toggle():

$('#menu').click(function() {
    $('.undermenu').toggle();
});

Though the reason your if($('.undermenu').css('display','block')) didn't work is because you set the display property of the element(s) to block, rather than getting the display property and testing it, which would be:

if ($('.undermenu').css('display') =='block')

If you really want to use an if to test the current display, before modifying the presentation, you'd have to do it inside of the click handler (otherwise it will only run once, on DOMReady, rather than every time):

$('#menu').click(function(){
    if ($('.undermenu').css('display') == 'block') {
        $('.undermenu').hide();
    }
    else {
        $('.undermenu').show();
    }
});

Or, if you want to risk the wrath of your colleagues, you can jazz that up a little:

$('#menu').click(function(){
    var undermenu = $('.undermenu');
    undermenu[undermenu.css('display') == 'block' ? 'hide' : 'show']();
});

References:

• css().
• toggle().

You use a setter on your condition :

// This line update .undermenu to display it and return true (return $('.undermenu'), so true)
if($('.undermenu').css('display','block')){

But you must get the value, and test

if($('.undermenu').css('display') === 'block'){

And you code conception is bad. If you do that, you test on document are ready if the .undermenu are displayed or not, and you put a function to the click trigger (to display or to hide..) but ! when your .undermenu was change, you already have the same trigger (to display or hide, and he never change)..

So you need to put your trigger for each click and test the value (displayed or not) on the trigger :

$(document).ready(function(){
    $('#menu').click(function(){
        if($('.undermenu').css('display') === 'block'){
           $('.undermenu').hide();
        }
        else {
           $('.undermenu').show();
        }
    });
});

On jquery exists:

$("#element_id").show();

$("#element_id").hide();

also you can use:

$("#element_id").fadeIn();

$("#element_id").fadeOut();

This show and hide elements with fade effects.

You can query if the element is hidden:

if($("#element_id").is(":hidden")){
    $("#element_id").show():
}
else{
    $("#element_id").hide();
}

What you're looking for is .toggle()

$("div").click(function () {
    $("img").toggle("slow", function () {
    // Animation plete.
    });
});

Here is a fiddle: http://jsfiddle/FQj89/

You can put the if statement in the function so you don't repeat yourself. You can probably also use toggle(); type stuff depending on what you are doing. ---

$('#menu').click(function(){

    if ( $('.undermenu').is(':visible') ) {

        $('.undermenu').hide();

     else {

        $('.undermenu').show();

    }

});

This is a good way too, depending on what you are doing though, one may be better than the other.

if ( $('.undermenu').css('display') === 'block' ) { // do something }