Question

使用下面的代码，用jade编写，我想我只能通过引用它的ID然后设置一个新样式来改变div id =“content”。但是，当我在浏览器中查看运行此代码的结果时，内容是可见的而不是隐藏的。有人可以让我深入了解我做错了什么，以及如何正确实现隐藏这样的元素？

   extends layout

    block content
      #content
        h5 good content test

    script.
        var searchVis = "visibility:" + search || "visibility:hidden";
        var uploadVis = "visibility:" + upload || "visibility:hidden";
        document.getElementById("content").style.visibility = "hidden";

Answer 1

当我刚刚尝试你的玉标记时，脚本失败了，因为你没有声明search或upload变量。如果您在浏览器中使用F12开发工具，则可以看到此错误。

以下对Jade 1.3.1来说效果很好：

extends layout

block content
  #content
    h5 good content test

  script.
    var search="";var upload="";
    var searchVis = "visibility:" + search || "visibility:hidden";
    var uploadVis = "visibility:" + upload || "visibility:hidden";
    document.getElementById("content").style.visibility = "hidden";

使用Jade html语言更改样式属性？

1 个答案: