jquery - jQuery选择器和CSS错误

问题1：我有三个变量。 className引用一个字符串。返回包含内容，链接包含
html字符串。我试图将内容分配给任何变量className引用。

var back = "test";
var className = "link1";
var link1 = '<div id="back" class="gohome"></div><div class="link3"></div><div class="link4"></div>';

因为以下代码似乎不适用于我

$(className) = className + back;

问题2：以下代码似乎并未更新prevClass包含的变量的CSS。

       $('[id*="back"]').each(function(){
            $('.' + prevClass).css({'left': '-123px'});
            $('.' + prevClass).css({'top': '430px'});
            $('.' + prevClass).css({'width': '123px'});
            $('.' + prevClass).css({'height': '44px'});

<script type="text/javascript">

    var back = "";
    var prevClass = "start"
    var className = "Broken";

    start = '<div id="back" class="gohome"></div><div class="link1"></div><div class="link2"></div>';
    gohome = '<div class="gohome"></div><div class="start"></div>';
    link1 = '<div id="back" class="gohome"></div><div class="link3"></div><div class="link4"></div>';
    link2 = '<div class="gohome"></div>';
    link3 = '<div id="back" class="gohome"></div><div class="link5"></div><div class="link6"></div>';
    link4 = '<div class="gohome"></div><div class="link7"></div><div class="link8"></div>';
    link5 = '<div class="gohome"></div><div class="link2a"></div><div class="link10"></div>';
    link6 = '<div class="gohome"></div><div class="link7"></div><div class="link11"></div>';
    link6a = '<div class="gohome"></div><div class="link7"></div><div class="link11"></div>';
    link7 = '<div class="gohome"></div><div class="link6a"></div><div class="link9"></div>';
    link8 = '<div class="gohome"></div><div class="link2a"></div><div class="link10"></div>';
    link9 = '<div class="gohome"></div><div class="link10a"></div>';
    link10 = '<div class="gohome"></div><div class="link12"></div><div class="link13"></div>';
    link10a = '<div class="gohome"></div><div class="link12"></div><div class="link13"></div>';
    link11 = '<div class="gohome"></div><div class="link10a"></div>';
    link12 = '<div class="gohome"></div><div class="link12"></div><div class="link13"></div>';
    link13 = '<div class="gohome"></div><div class="link14"></div><div class="link15"></div>';
    link14 = '<div class="gohome"></div>';
    link15 = '<div class="gohome"></div><div class="link16"></div><div class="link17"></div>';
    link16 = '<div class="gohome"></div>';
    link17 = '<div class="gohome"></div><div class="link18"></div><div class="link19"></div>';
    link18 = '<div class="gohome"></div>';
    link19 = '<div class="gohome"></div><div class="link20"></div><div class="link21"></div>';
    link20 = '<div class="gohome"></div>';
    link21 = '<div class="gohome"></div>';



   $(document).on('click', '.inter [class]', function () {
   className = this.className;
   $('[id*="back"]').each(function(){
        $('.' + prevClass).css({'left': '123px'});
        $('.' + prevClass).css({'top': '430px'});
        $('.' + prevClass).css({'width': '123px'});
        $('.' + prevClass).css({'height': '44px'});
    });

   back = '<div class="' + prevClass + '"></div>';
   $('.' + className).html(back);
   prevClass = className;
       $('.inter').fadeTo(250, 0.25, function () {
           $('.inter').html(window[className]);

           $('.inter').css({'background-image': 'url("' + className + '.png")'});
           $('.inter').fadeTo(250, 1.00);

       });
   });

</script>

最佳答案

对于问题一，您需要选择和更新一点点。由于classname是一个类名，因此您需要：

$('.' + className)

（就像您已经找出问题2一样）

如果要更新内容（带有背面）：

$('.' + className).html(back);

对于问题2，目前尚不清楚您要做什么。

首先，#back已经在DOM中了吗？这对您的link1变量无效。第二，prevClass应该是什么？

关于jquery - jQuery选择器和CSS错误，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/16682775/