Swapping out DIVs with jQuery

Question

Swapping out DIVs with jQuery

Currently, I am working on a project for a client where I need to create a sortable biography section using jQuery.

Initially, I had a setup with just two bios (first person and third person) which was functioning perfectly:

$("a#first").click(function(){
    $("div.third").fadeOut("",function(){
        $("div.first").fadeIn("");
    });
});

You can view the original setup in this fiddle:

http://jsfiddle.net/Vp7BL/1/

However, we have now decided to change course and include three different bio lengths (short, medium, long). I attempted to modify the code by including the extra div into the selector as shown below (line 2), but it is not functioning smoothly and seems to be lagging:

$("a#short").click(function(){
        $("div.medium, div.long").fadeOut("",function(){
        $("div.short").fadeIn("");
    }); 
});

You can check out the current setup in this fiddle:

http://jsfiddle.net/Vp7BL/

I would greatly appreciate any assistance or guidance on how to resolve this issue. Thank you!

jquery css fadein fadeout

Answer 1

Answer №1

Take a look at this revision I made to your code. I recommend using absolute positioning for the bios to ensure they all align vertically from the same starting point, eliminating any noticeable 'jump' during transitions.

.short, .medium, .long {
    position: absolute;
    top: 40px;
}

Answer 2

Take a look at this revision I made to your code. I recommend using absolute positioning for the bios to ensure they all align vertically from the same starting point, eliminating any noticeable 'jump' during transitions.

.short, .medium, .long {
    position: absolute;
    top: 40px;
}

Answer 3

Answer №2

To make things simpler, consider adding an "active" class to the current active bio. This will help with managing your code more effectively.

$('a#short').click(function() {
  $('div.active').removeClass("active").fadeOut("", function() {
    $('div.short').addClass("active").fadeIn("");
  });
});

You could enhance the code further by making it more modular. This way, you won't have to update the JavaScript every time you add or remove a bio.

Imagine if your links were structured like this:

<ul id="bionav">
  <a class="bio-link" data-bio="short">short</a>
  <a class="bio-link" data-bio="med">med</a>
  <a class="bio-link" data-bio="long">long</a>
</ul>
<ul id="biodisplay">
  <div class="bio-link short">short</div>
  <div class="bio-link med">med</div>
  <div class="bio-link long">long</div>
</ul>

You can manage your set effectively with code like this:

$('#bionav').on('click', '.bio-link', function() {
  var $bioDisplay = $('#biodisplay'),
  $targetBio = $(this).attr('data-bio'); //if ie doesn't matter you can do .data('bio');
  $active = $bioDisplay.find('.active');

  if ($active != undefined) {
    $active.removeClass('active').fadeOut("", showBio($targetBio));
  } else {
    showBio($targetBio);
  }
});

function showBio(bio) {
  bio.addClass('active').fadeIn("");
}

This approach also incorporates event delegation, allowing you to attach just one event to the DOM instead of three.

Considering Josh's suggestion, you don't necessarily have to make the fadeIn a callback to the fadeOut. You can initiate them simultaneously for a smoother crossfade effect.

Answer 4

To make things simpler, consider adding an "active" class to the current active bio. This will help with managing your code more effectively.

$('a#short').click(function() {
  $('div.active').removeClass("active").fadeOut("", function() {
    $('div.short').addClass("active").fadeIn("");
  });
});

You could enhance the code further by making it more modular. This way, you won't have to update the JavaScript every time you add or remove a bio.

Imagine if your links were structured like this:

<ul id="bionav">
  <a class="bio-link" data-bio="short">short</a>
  <a class="bio-link" data-bio="med">med</a>
  <a class="bio-link" data-bio="long">long</a>
</ul>
<ul id="biodisplay">
  <div class="bio-link short">short</div>
  <div class="bio-link med">med</div>
  <div class="bio-link long">long</div>
</ul>

You can manage your set effectively with code like this:

$('#bionav').on('click', '.bio-link', function() {
  var $bioDisplay = $('#biodisplay'),
  $targetBio = $(this).attr('data-bio'); //if ie doesn't matter you can do .data('bio');
  $active = $bioDisplay.find('.active');

  if ($active != undefined) {
    $active.removeClass('active').fadeOut("", showBio($targetBio));
  } else {
    showBio($targetBio);
  }
});

function showBio(bio) {
  bio.addClass('active').fadeIn("");
}

This approach also incorporates event delegation, allowing you to attach just one event to the DOM instead of three.

Considering Josh's suggestion, you don't necessarily have to make the fadeIn a callback to the fadeOut. You can initiate them simultaneously for a smoother crossfade effect.

Answer 5

Answer №3

Is it possible to have two divs visible simultaneously? If not, then why are you fading out two of them? The correct approach would be as follows: 1. Assign a class (not in the current location) to the currently displayed div, let's call it "selected". 2. When a tab is clicked, fade out the element with the "selected" class. In the callback function for the fadeout, remove the "selected" class from the previously displayed element and add it to the newly clicked tab.

Alternatively, consider utilizing jquery-ui for additional functionality. Best of luck!

Answer 6

Is it possible to have two divs visible simultaneously? If not, then why are you fading out two of them? The correct approach would be as follows: 1. Assign a class (not in the current location) to the currently displayed div, let's call it "selected". 2. When a tab is clicked, fade out the element with the "selected" class. In the callback function for the fadeout, remove the "selected" class from the previously displayed element and add it to the newly clicked tab.

Alternatively, consider utilizing jquery-ui for additional functionality. Best of luck!

Answer 7

Answer №4

Here's a suggestion for improving your "design". By adding Josh Rutherford's position: absolute CSS style, you can maintain the same markup structure while enhancing manageability and resolving any existing issues.

Check out the CSS:

.short, .medium, .long {
    position: absolute;
    top: 30px; /* adjust as needed */
}

And here's some JavaScript:

$("a#short, a#medium, a#long").on("click", function() {
    var elid = $(this).attr('id');
    // Call getTab based on ID
    getTab(elid);
});

function getTab(tab, speed=500) {
  // Get current tab
  var the_tab = $("#" + tab);

  // Find corresponding content div for current tab
  var content = $("."+tab);

  // Store sibling tabs' classes in an array
  sibs = [];

  the_tab.siblings().each(function () {
      sibs.push('.' + $(this).attr('id'));
      // Deselect siblings
      $(this).removeClass('selected');
  });

  // Fade out siblings
  $(sibs.join(",")).fadeOut(speed, function() {
      // Fade in current tab and mark as selected
      the_tab.addClass('selected');
      content.fadeIn(speed);
  });
}

Answer 8

Here's a suggestion for improving your "design". By adding Josh Rutherford's position: absolute CSS style, you can maintain the same markup structure while enhancing manageability and resolving any existing issues.

Check out the CSS:

.short, .medium, .long {
    position: absolute;
    top: 30px; /* adjust as needed */
}

And here's some JavaScript:

$("a#short, a#medium, a#long").on("click", function() {
    var elid = $(this).attr('id');
    // Call getTab based on ID
    getTab(elid);
});

function getTab(tab, speed=500) {
  // Get current tab
  var the_tab = $("#" + tab);

  // Find corresponding content div for current tab
  var content = $("."+tab);

  // Store sibling tabs' classes in an array
  sibs = [];

  the_tab.siblings().each(function () {
      sibs.push('.' + $(this).attr('id'));
      // Deselect siblings
      $(this).removeClass('selected');
  });

  // Fade out siblings
  $(sibs.join(",")).fadeOut(speed, function() {
      // Fade in current tab and mark as selected
      the_tab.addClass('selected');
      content.fadeIn(speed);
  });
}

Answer 9

Answer №5

Let's get started:

The primary issue lies in the need to use absolute positioning for the .short, .medium and .long classes to avoid any layout problems. Additionally, consider nesting them inside another class for easier jQuery selection if you plan to add an .extralong class in the future.

With a few adjustments, your code should appear similar to this:

Jquery:

$('.toggles a').click(function(e) {
    //Ensure to include preventDefault
    e.preventDefault();
    //If already selected, do not proceed
    if($(this).hasClass('selected'))
        return false;
    //Fade out all text elements 
    $('.short, .medium, .long').fadeOut('10');
    //Fade in the clicked element
    $('.'+$(this).attr('id')).fadeIn();
    //Remove 'selected' class from all elements
    $('.toggles a').removeClass('selected');
    //Add 'selected' class to clicked element
    $(this).addClass('selected');
});

CSS:

.short, .medium, .long{
    position: absolute;
}

Fiddle:

http://jsfiddle.net/Vp7BL/3/

Answer 10

Let's get started:

The primary issue lies in the need to use absolute positioning for the .short, .medium and .long classes to avoid any layout problems. Additionally, consider nesting them inside another class for easier jQuery selection if you plan to add an .extralong class in the future.

With a few adjustments, your code should appear similar to this:

Jquery:

$('.toggles a').click(function(e) {
    //Ensure to include preventDefault
    e.preventDefault();
    //If already selected, do not proceed
    if($(this).hasClass('selected'))
        return false;
    //Fade out all text elements 
    $('.short, .medium, .long').fadeOut('10');
    //Fade in the clicked element
    $('.'+$(this).attr('id')).fadeIn();
    //Remove 'selected' class from all elements
    $('.toggles a').removeClass('selected');
    //Add 'selected' class to clicked element
    $(this).addClass('selected');
});

CSS:

.short, .medium, .long{
    position: absolute;
}

Fiddle:

http://jsfiddle.net/Vp7BL/3/

Swapping out DIVs with jQuery

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Jquery:

CSS:

Fiddle:

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