使用jQuery在单击时显示/隐藏div

Question

我有4个Divs，并且单击导航，我想显示其中一个并隐藏其他人。我的工作正常，但我觉得它不像我知道的那样光滑，它绝对需要重新制作我的代码！我拥有的是我的。

$('#details-speakers').click(function() {

    $('#home').slideUp('slow', function() {});
    $('#sessions-content').slideUp('slow', function() {});
    $('#cases-content').slideUp('slow', function() {});
    $('html,body').animate({scrollTop: $("#details").offset().top - 16}, 200, "swing");
  $('#speakers-content').slideDown('slow', function() {
        $('#details-speakers').addClass('selected');
        //Remove other classes
        $('#details-sessions').removeClass('selected');
        $('#details-cases').removeClass('selected');
        $('#details-workshops').removeClass('selected');
  });
});
$('#details-sessions').click(function() {

    $('#home').slideUp('slow', function() {});
    $('#speakers-content').slideUp('slow', function() {});
    $('#cases-content').slideUp('slow', function() {});
    $('html,body').animate({scrollTop: $("#details").offset().top - 16}, 200, "swing");
  $('#sessions-content').slideDown('slow', function() {
    $('#details-sessions').addClass('selected');
        //Remove other classes
        $('#details-speakers').removeClass('selected');
        $('#details-cases').removeClass('selected');
        $('#details-workshops').removeClass('selected');
  });
});
$('#details-cases').click(function() {

    $('#home').slideUp('slow', function() {});
    $('#speakers-content').slideUp('slow', function() {});
    $('#sessions-content').slideUp('slow', function() {});
    $('html,body').animate({scrollTop: $("#details").offset().top - 16}, 200, "swing");
  $('#cases-content').slideDown('slow', function() {
    $('#details-cases').addClass('selected');
        //Remove other classes
        $('#details-speakers').removeClass('selected');
        $('#details-sessions').removeClass('selected');
        $('#details-workshops').removeClass('selected');
  });
});

Answer 1

平滑度或缺乏并不是您的代码的错。一些浏览器很慢。这不是可以从JS固定的东西。

您可以使代码较短，但我认为不会进行任何物质性能的改进。

$('#details-speakers').click(function() {
    $('#home, #sessions-content, #cases-content').slideUp('slow');
    $('body').animate({scrollTop: $("#details").offset().top - 16}, 200, "swing");
  $('#speakers-content').slideDown('slow', function() {
        $('#details-speakers').addClass('selected');
        //Remove other classes
        $('#details-sessions, #details-cases, #details-workshops').removeClass('selected');
  });
});

$('#details-sessions').click(function() {
    $('#home, #speakers-content, #cases-content').slideUp('slow');
    $('body').animate({scrollTop: $("#details").offset().top - 16}, 200, "swing");
  $('#sessions-content').slideDown('slow', function() {
    $('#details-sessions').addClass('selected');
        //Remove other classes
        $('#details-speakers, #details-cases, #details-workshops').removeClass('selected');
  });
});

$('#details-cases').click(function() {
    $('#home, #speakers-content, #sessions-content').slideUp('slow');
    $('body').animate({scrollTop: $("#details").offset().top - 16}, 200, "swing");
  $('#cases-content').slideDown('slow', function() {
    $('#details-cases').addClass('selected');
        //Remove other classes
        $('#details-speakers, #details-sessions, #details-workshops').removeClass('selected');
  });
});

Answer 2

Ctrl+C和Ctrl+V编程直接导致地狱...我认为您可以做这样的事情：

var divs = $('#details-speakers, #details-sessions, #details-cases');

divs.click(function () {
    divs.not(this).add('#home').slideUp('slow');
    // animation with home and body
    $(this).slideDown('slow', function () {
        $(this).addClass('selected');
        divs.not(this).removeClass('selected');
    });
}

性能取决于很多事情（Firebug启动，Divs的内容，Divs'样式，半透明背景等） - 不仅仅是JavaScript。

Answer 3

如果这是一种更清洁的方法，您正在寻找这样的事情：

$(".my4Divs").click(function() {
    var self = this;
    $(".my4Divs").each(function() {
       if(self == this) {
          $(self).addClass("selected");
          ...
          ...
          return;
       }
       $(this).removeClass("selected");
       ...
       ...
    });
});